Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration) Questions in English

(C) In uniform circular motion,the speed of the particle remains constant,but the direction of velocity changes continuously at every point along the path.
Since acceleration is defined as the rate of change of velocity,this change in direction results in an acceleration.
This acceleration is directed towards the center of the circular path and is known as centripetal or radial acceleration.
Therefore,the acceleration is an internal radial acceleration directed towards the center.

(A) For a body to move in a uniform circular motion,it must constantly change its direction of velocity.
According to Newton's first law,a body requires an external force to change its state of motion (direction).
This force,which acts towards the center of the circular path and is perpendicular to the velocity vector,is known as the centripetal force.
Therefore,the correct answer is $A$.

(B) For a body of mass $m$ moving in a circle of radius $r$ with a constant speed $v$,the acceleration directed towards the center is known as centripetal acceleration,given by $a_c = v^2/r$.
According to Newton's second law of motion,the force $F$ is the product of mass and acceleration $(F = ma)$.
Therefore,the magnitude of the centripetal force is $F_c = m \cdot a_c = m(v^2/r) = mv^2/r$.
Thus,the correct option is $B$.

(A) Centripetal acceleration $(a_c)$ is the acceleration of an object moving in a circular path directed towards the center of the circle.
For an object moving with a constant speed $(v)$ in a circular path of radius $(r)$,the magnitude of the centripetal acceleration is given by the formula:
$a_c = \frac{v^2}{r}$
Therefore,the correct option is $(A)$.

(A) Given: Radius $r = 50\, cm$,Number of revolutions $N = 10$,Time $t = 20\, s$.
Frequency $n = \frac{N}{t} = \frac{10}{20} = 0.5\, Hz = \frac{1}{2}\, Hz$.
The centripetal acceleration $a$ is given by $a = \omega^2 r = (2\pi n)^2 r = 4\pi^2 n^2 r$.
Substituting the values: $a = 4 \times (3.14)^2 \times (0.5)^2 \times 50$.
$a = 4 \times 9.8596 \times 0.25 \times 50$.
$a = 492.98\, cm/s^2 \approx 493\, cm/s^2$.

(A) The acceleration of an object moving in a circular path is the centripetal acceleration,given by the formula $a = \frac{v^2}{r}$.
Given,speed $v = 400 \, m/s$ and radius $r = 160 \, m$.
Substituting the values into the formula:
$a = \frac{(400)^2}{160} = \frac{160000}{160} = 1000 \, m/s^2$.
Since $1000 \, m/s^2 = 1 \, km/s^2$,the correct option is $A$.

(A) The formula for centripetal acceleration $a$ is given by $a = v\omega$,where $v$ is the orbital speed and $\omega$ is the angular velocity.
Given that the new orbital speed $v' = 2v$ and the new angular velocity $\omega' = \frac{\omega}{2}$.
The new centripetal acceleration $a'$ is calculated as:
$a' = v' \times \omega' = (2v) \times \left( \frac{\omega}{2} \right) = v\omega = a$.
Therefore,the centripetal acceleration remains unchanged.

(B) The formula for centripetal force is $F = \frac{mv^2}{R}$.
Given that the mass $m$ and speed $v$ remain constant,the centripetal force $F$ is inversely proportional to the radius $R$,i.e.,$F \propto \frac{1}{R}$.
If the radius is doubled $(R' = 2R)$,the new force $F'$ becomes $F' = \frac{mv^2}{2R} = \frac{1}{2} F$.
Therefore,the centripetal force is halved.

(B) The car is undergoing non-uniform circular motion.
There are two components of acceleration:
$1$. Tangential acceleration $(a_t)$: This is the rate of change of speed,given as $a_t = 2 \ m/s^2$.
$2$. Centripetal acceleration $(a_c)$: This is directed towards the center of the circular path,given by $a_c = \frac{v^2}{r}$.
Substituting the values: $a_c = \frac{(30)^2}{500} = \frac{900}{500} = 1.8 \ m/s^2$.
The net acceleration $(a)$ is the vector sum of these two perpendicular components:
$a = \sqrt{a_t^2 + a_c^2} = \sqrt{(2)^2 + (1.8)^2} = \sqrt{4 + 3.24} = \sqrt{7.24} \approx 2.69 \ m/s^2$.
Rounding to one decimal place,the acceleration is $2.7 \ m/s^2$.

(D) In circular motion,the velocity vector is always tangent to the circular path,which is referred to as the transverse direction.
In non-uniform circular motion,the particle experiences two components of acceleration:
$1$. Centripetal (radial) acceleration $(a_r = v^2/r)$,which changes the direction of velocity.
$2$. Tangential (transverse) acceleration $(a_t = dv/dt)$,which changes the magnitude of velocity.
Since both components are present,the acceleration has both radial and transverse components,while the velocity remains transverse.

(D) The centripetal acceleration $(a_c = v^2/r)$ is required for any circular motion to change the direction of the velocity vector.
Tangential acceleration $(a_t = dv/dt)$ is responsible for changing the magnitude of the velocity.
Since $a_c$ depends on the speed and radius,and $a_t$ depends on the rate of change of speed,there is no fixed relationship between them.
In uniform circular motion,$a_t = 0$ while $a_c \neq 0$.
In non-uniform circular motion,both can have any arbitrary values depending on the kinematics of the particle.
Therefore,the centripetal acceleration may be more,less,or even equal to the tangential acceleration depending on the specific motion.

(D) $1$. In circular motion,the angular momentum is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
$2$. For a particle moving in a circular path,the position vector $\vec{r}$ and the velocity vector $\vec{v}$ lie in the plane of the circle.
$3$. The angular momentum vector $\vec{L}$ is perpendicular to the plane of the circle (along the axis of rotation) according to the right-hand rule.
$4$. Even if the speed $v$ is decreasing,the direction of the axis of rotation remains fixed perpendicular to the plane of the circular path.
$5$. Therefore,the direction of the angular momentum vector remains constant,although its magnitude changes as the speed changes.

(D) When a particle moves in a circular path,its position vector $\vec{r}$ and velocity vector $\vec{v}$ lie in the plane of the circle.
The angular momentum is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
Since the motion is confined to a fixed plane (the plane of the circle),the cross product $\vec{r} \times \vec{v}$ always points in a direction perpendicular to the plane of the circle according to the right-hand rule.
Even if the speed of the particle changes,the direction of the angular momentum vector remains constant (perpendicular to the plane of motion).
Therefore,the correct statement is that the direction of angular momentum remains constant.

(D) The resultant acceleration $a$ is given by $a = \sqrt{a_T^2 + a_r^2}$.
Substituting the given values: $a = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \ m/s^2$.
From the geometry of the vector triangle,the radial acceleration $a_r$ is the adjacent side and the resultant acceleration $a$ is the hypotenuse for the angle $\theta$.
Therefore,$\cos \theta = \frac{a_r}{a} = \frac{3}{5}$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we have $\sec \theta = \frac{5}{3}$.

(B) The formula for centripetal force is $F = \frac{mv^2}{r}$.
Given that mass $m$,speed $v$,and radius $r$ are increased by $50\%$,the new values are $m' = 1.5m$,$v' = 1.5v$,and $r' = 1.5r$.
The new centripetal force $F'$ is given by:
$F' = \frac{m' (v')^2}{r'} = \frac{(1.5m)(1.5v)^2}{1.5r} = \frac{1.5 \times 2.25}{1.5} \frac{mv^2}{r} = 2.25 F$.
The percentage increase in force is given by:
$\frac{F' - F}{F} \times 100\% = \frac{2.25F - F}{F} \times 100\% = 1.25 \times 100\% = 125\%$.

(B) The centripetal acceleration is given by $a_c = \frac{v^2}{r} = k^2 r t^2$.
From this,we find the velocity $v$:
$v^2 = k^2 r^2 t^2 \implies v = k r t$.
The tangential acceleration $a_t$ is the rate of change of speed:
$a_t = \frac{dv}{dt} = \frac{d}{dt}(k r t) = k r$.
The power $P$ delivered by the force is given by $P = F_t v$,where $F_t = m a_t$ is the tangential force.
$P = (m a_t) v = m(k r)(k r t) = m k^2 r^2 t$.

(C) Given,velocity $v = 1.0t$ and radius $r = 0.1 \ m$.
Tangential acceleration $a_t = \frac{dv}{dt} = \frac{d}{dt}(1.0t) = 1.0 \ m/s^2$.
At $t = 5 \ s$,the velocity $v = 1.0 \times 5 = 5 \ m/s$.
Centripetal acceleration $a_c = \frac{v^2}{r} = \frac{5^2}{0.1} = \frac{25}{0.1} = 250 \ m/s^2$.
The total acceleration $a_{net} = \sqrt{a_c^2 + a_t^2} = \sqrt{250^2 + 1^2} = \sqrt{62500 + 1} = \sqrt{62501} \approx 250 \ m/s^2$.

(C) The car is moving in a circular path,so it experiences two types of acceleration:
$1$. Centripetal acceleration $(a_c)$,which is directed towards the center: $a_c = \frac{v^2}{r}$.
$2$. Tangential acceleration $(a_t)$,which is given as $a_t = g$.
Since these two accelerations are perpendicular to each other,the net acceleration $(a_{net})$ is the vector sum of the two:
$a_{net} = \sqrt{a_c^2 + a_t^2}$
Substituting the values:
$a_{net} = \sqrt{(\frac{v^2}{r})^2 + g^2}$
$a_{net} = \sqrt{\frac{v^4}{r^2} + g^2} = [\frac{v^4}{r^2} + g^2]^{1/2}$.

(B) Given: Tangential acceleration $a_t = 2 \, m/s^2$,velocity $v = 30 \, m/s$,and radius $r = 500 \, m$.
First,calculate the centripetal acceleration $a_c$ using the formula $a_c = \frac{v^2}{r}$.
$a_c = \frac{30^2}{500} = \frac{900}{500} = 1.8 \, m/s^2$.
The total acceleration $a$ is the vector sum of tangential and centripetal acceleration,given by $a = \sqrt{a_t^2 + a_c^2}$.
$a = \sqrt{2^2 + 1.8^2} = \sqrt{4 + 3.24} = \sqrt{7.24}$.
$a \approx 2.69 \, m/s^2$,which rounds to $2.7 \, m/s^2$.

(B) Given displacement $S = \frac{t^2}{2} + \frac{t^3}{3}$.
Velocity $v = \frac{dS}{dt} = t + t^2$.
At $t = 2 \, s$,$v = 2 + (2)^2 = 6 \, m/s$.
Tangential acceleration $a_t = \frac{dv}{dt} = 1 + 2t$.
At $t = 2 \, s$,$a_t = 1 + 2(2) = 5 \, m/s^2$.
Centripetal acceleration $a_c = \frac{v^2}{r} = \frac{6^2}{3} = \frac{36}{3} = 12 \, m/s^2$.
The total acceleration $a = \sqrt{a_t^2 + a_c^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, m/s^2$.

(A) Given:
Total acceleration $a = 15 \, m s^{-2}$
Radius $R = 2.5 \, m$
Angle between total acceleration and centripetal acceleration is $30^{\circ}$.
The centripetal acceleration $a_c$ is the component of the total acceleration $a$ directed towards the center of the circle.
From the figure,$a_c = a \cos(30^{\circ})$.
$a_c = 15 \times \frac{\sqrt{3}}{2} \approx 15 \times 0.866 = 12.99 \, m s^{-2}$.
We know that the centripetal acceleration is given by $a_c = \frac{v^2}{R}$,where $v$ is the speed of the particle.
Therefore,$v = \sqrt{a_c R}$.
$v = \sqrt{12.99 \times 2.5} = \sqrt{32.475} \approx 5.698 \, m/s$.
Rounding to one decimal place,we get $v \approx 5.7 \, m/s$.
Thus,the correct option is $A$.

(D) Given: Radius $r = 50\, cm = 0.5\, m$,angular acceleration $\alpha = 2.0\, rad\, s^{-2}$,initial angular velocity $\omega_0 = 0$,and time $t = 2.0\, s$.
First,calculate the angular velocity $\omega$ at $t = 2.0\, s$ using $\omega = \omega_0 + \alpha t = 0 + (2.0)(2.0) = 4.0\, rad\, s^{-1}$.
The tangential acceleration is $a_t = r\alpha = 0.5 \times 2.0 = 1.0\, m\, s^{-2}$.
The radial (centripetal) acceleration is $a_r = \omega^2 r = (4.0)^2 \times 0.5 = 16 \times 0.5 = 8.0\, m\, s^{-2}$.
The net acceleration $a$ is given by $a = \sqrt{a_t^2 + a_r^2} = \sqrt{1.0^2 + 8.0^2} = \sqrt{1 + 64} = \sqrt{65} \approx 8.06\, m\, s^{-2}$.
Rounding to the nearest value,the net acceleration is approximately $8.0\, m\, s^{-2}$.

(B) In circular motion,the car experiences two types of acceleration:
$1$. Radial (centripetal) acceleration: $a_r = \frac{v^2}{r}$
$2$. Tangential acceleration: $a_t = a$ (given as the rate of change of speed)
Since these two accelerations are perpendicular to each other,the resultant acceleration $a_{net}$ is given by:
$a_{net} = \sqrt{a_r^2 + a_t^2}$
Substituting the values:
$a_{net} = \sqrt{\left(\frac{v^2}{r}\right)^2 + a^2} = \sqrt{\frac{v^4}{r^2} + a^2}$

(C) From the given graph,the tangential acceleration $a_T$ is a linear function of time $t$ passing through the origin. The slope of the line is $\tan 60^{\circ} = \sqrt{3}$.
Thus,$a_T = \sqrt{3} t$.
Since $a_T = \frac{dv}{dt}$,we have $\frac{dv}{dt} = \sqrt{3} t$.
Integrating with respect to time with initial velocity $v(0) = 0$,we get $v = \int_0^t \sqrt{3} t \ dt = \frac{\sqrt{3} t^2}{2}$.
The centripetal (radial) acceleration is $a_c = \frac{v^2}{r}$. Given $r = 1 \ m$,$a_c = \frac{(\sqrt{3} t^2 / 2)^2}{1} = \frac{3 t^4}{4}$.
The total acceleration vector $\vec{a}$ makes an angle $\theta = 30^{\circ}$ with the radial acceleration vector $\vec{a}_c$. Since $\vec{a}_c$ and $\vec{a}_T$ are perpendicular,we have $\tan \theta = \frac{a_T}{a_c}$.
Substituting the values: $\tan 30^{\circ} = \frac{\sqrt{3} t}{3 t^4 / 4} \Rightarrow \frac{1}{\sqrt{3}} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{t^3}$.
Solving for $t$: $t^3 = 4 \sqrt{3} \times \sqrt{3} = 4 \times 3 = 12$. This seems to contradict the options. Let's re-evaluate: $\tan 30^{\circ} = \frac{a_T}{a_c} \Rightarrow \frac{1}{\sqrt{3}} = \frac{\sqrt{3} t}{3 t^4 / 4} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{\sqrt{3} t^3}$.
Thus,$t^3 = 4 \times 3 / 1 = 12$ is incorrect. Let's re-check: $\tan 30^{\circ} = \frac{a_T}{a_c} = \frac{\sqrt{3} t}{3 t^4 / 4} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{\sqrt{3} t^3}$.
Wait,$\frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3} t^3} \Rightarrow t^3 = 4 \Rightarrow t = 4^{1/3} = 2^{2/3} \ s$.

(C) From the given graph,the tangential acceleration $a_T$ is a linear function of time $t$ passing through the origin with a slope of $\tan 60^{\circ} = \sqrt{3}$.
Thus,$a_T = \sqrt{3} t$.
Since $a_T = \frac{dv}{dt}$,we have $\frac{dv}{dt} = \sqrt{3} t$.
Integrating with respect to time (with initial velocity $v=0$ at $t=0$): $v = \int_{0}^{t} \sqrt{3} t \, dt = \frac{\sqrt{3}}{2} t^2$.
The centripetal (radial) acceleration is $a_c = \frac{v^2}{r}$. Given $r = 1 \, m$,we have $a_c = \frac{(\frac{\sqrt{3}}{2} t^2)^2}{1} = \frac{3}{4} t^4$.
The total acceleration vector $\vec{a}$ is the vector sum of the radial acceleration $\vec{a}_c$ and the tangential acceleration $\vec{a}_T$. Since $\vec{a}_c$ and $\vec{a}_T$ are perpendicular,the angle $\theta$ that the total acceleration makes with the radial acceleration $\vec{a}_c$ is given by $\tan \theta = \frac{a_T}{a_c}$.
We are given $\theta = 30^{\circ}$,so $\tan 30^{\circ} = \frac{a_T}{a_c}$.
$\frac{1}{\sqrt{3}} = \frac{\sqrt{3} t}{\frac{3}{4} t^4} = \frac{4 \sqrt{3} t}{3 t^4} = \frac{4}{\sqrt{3} t^3}$.
$t^3 = \frac{4 \times \sqrt{3}}{\sqrt{3}} = 4$.
$t = 4^{1/3} = (2^2)^{1/3} = 2^{2/3} \, s$.

(D) Given that the magnitude of radial acceleration $(a_r = v^2/R)$ and tangential acceleration $(a_t = dv/dt)$ are equal:
$dv/dt = v^2/R$
Rearranging the terms for integration:
$\int_{V_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{R}$
$[-1/v]_{V_0}^{v} = t/R \Rightarrow 1/V_0 - 1/v = t/R \Rightarrow t = R(1/V_0 - 1/v) \dots(1)$
Also,$a_t = v(dv/ds) = v^2/R$,which implies $dv/v = ds/R$.
Integrating for one full revolution ($s = 0$ to $s = 2\pi R$):
$\int_{V_0}^{v} \frac{dv}{v} = \int_{0}^{2\pi R} \frac{ds}{R}$
$\ln(v/V_0) = 2\pi \Rightarrow v = V_0 e^{2\pi} \dots(2)$
Substituting $v$ from $(2)$ into $(1)$:
$t = R/V_0 (1 - 1/e^{2\pi}) = R/V_0 (1 - e^{-2\pi})$.

(A) Given,the path length $s = t^3 + 5$.
Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(t^3 + 5) = 3t^2 \ m/s$.
Tangential acceleration $a_t = \frac{dv}{dt} = \frac{d}{dt}(3t^2) = 6t \ m/s^2$.
Radial (centripetal) acceleration $a_c = \frac{v^2}{R} = \frac{(3t^2)^2}{R} = \frac{9t^4}{R} \ m/s^2$.
At $t = 2 \ s$:
$a_t = 6 \times 2 = 12 \ m/s^2$.
$a_c = \frac{9 \times (2)^4}{20} = \frac{9 \times 16}{20} = \frac{144}{20} = 7.2 \ m/s^2$.
The resultant acceleration $a = \sqrt{a_t^2 + a_c^2}$.
$a = \sqrt{(12)^2 + (7.2)^2} = \sqrt{144 + 51.84} = \sqrt{195.84} \approx 14 \ m/s^2$.

(C) In the rotating frame of the disc,the insect experiences two pseudo-forces: the centrifugal force $F_c = m\omega^2r$ directed radially outwards and the Coriolis force $F_{cor} = 2m(\vec{v} \times \vec{\omega})$ directed perpendicular to the radial velocity.
In the ground frame,the acceleration of the insect is the vector sum of radial and tangential components.
The radial acceleration is $a_r = -r\omega^2$ (centripetal acceleration).
The tangential acceleration is $a_t = 2v\omega$ (due to the change in tangential velocity as the insect moves radially).
The total acceleration magnitude is $a = \sqrt{a_r^2 + a_t^2} = \sqrt{(r\omega^2)^2 + (2v\omega)^2}$.
Since $(2v\omega)^2 > 0$,the magnitude $a = \sqrt{r^2\omega^4 + 4v^2\omega^2}$ is clearly greater than $r\omega^2$.

(B) The acceleration vector $\vec{a}$ can be resolved into tangential acceleration $\vec{a}_t$ and centripetal acceleration $\vec{a}_c$.
Tangential acceleration $\vec{a}_t$ is parallel to the velocity vector $\vec{v}$ (and thus parallel to momentum $\vec{p}$),while centripetal acceleration $\vec{a}_c$ is perpendicular to $\vec{v}$.
We calculate the dot product of $\vec{a}$ and $\vec{p}$ to determine the nature of the speed change:
$\vec{a} \cdot \vec{p} = (4\hat{i} + 3\hat{j}) \cdot (8\hat{i} - 6\hat{j}) = (4 \times 8) + (3 \times -6) = 32 - 18 = 14$.
Since $\vec{a} \cdot \vec{p} > 0$,the angle between $\vec{a}$ and $\vec{v}$ is acute.
This implies that the tangential component of acceleration is in the direction of velocity,causing the speed of the particle to increase.
Therefore,the motion is accelerated circular motion.

(B) Given,$v = a\sqrt{s}$.
Tangential acceleration $a_t = v \frac{dv}{ds}$.
Since $\frac{dv}{ds} = a \cdot \frac{1}{2\sqrt{s}} = \frac{a}{2\sqrt{s}}$,we have $a_t = (a\sqrt{s}) \cdot (\frac{a}{2\sqrt{s}}) = \frac{a^2}{2}$.
Centripetal acceleration $a_c = \frac{v^2}{R} = \frac{(a\sqrt{s})^2}{R} = \frac{a^2 s}{R}$.
The angle $\alpha$ between the total acceleration vector and the velocity vector is given by $\tan \alpha = \frac{a_c}{a_t}$.
Substituting the values,$\tan \alpha = \frac{a^2 s / R}{a^2 / 2} = \frac{2s}{R}$.

(B) In uniform circular motion,a body moves along a circular path with a constant speed.
Although the speed is constant,the direction of velocity changes at every point,which implies that the body is accelerating.
This acceleration is known as centripetal acceleration,which is directed towards the center of the circle.
According to Newton's second law of motion $(F = ma)$,the net force acting on the body must be in the same direction as the acceleration.
Therefore,the net force (centripetal force) is directed towards the center of the circle,which is always perpendicular to the instantaneous velocity (direction of motion).
Since the speed $(v)$ and the radius $(r)$ are constant,the magnitude of the centripetal force $(F = mv^2/r)$ remains constant.

(C) The tangential acceleration is given as $a_t = 2 \ m/s^2$.
The centripetal acceleration is given by the formula $a_c = \frac{v^2}{r}$.
Substituting the given values,$a_c = \frac{30^2}{500} = \frac{900}{500} = 1.8 \ m/s^2$.
The net acceleration $a_{net}$ is the vector sum of tangential and centripetal acceleration,given by $a_{net} = \sqrt{a_t^2 + a_c^2}$.
$a_{net} = \sqrt{2^2 + 1.8^2} = \sqrt{4 + 3.24} = \sqrt{7.24} \approx 2.69 \ m/s^2$.
Rounding to one decimal place,the net acceleration is $2.7 \ m/s^2$.

(B) Given kinetic energy $k = \frac{1}{2}mv^2 = as^2$.
From this,the velocity is $v = s\sqrt{\frac{2a}{m}}$.
The centripetal acceleration is $a_R = \frac{v^2}{R} = \frac{2as^2}{mR}$.
The tangential acceleration is $a_t = \frac{dv}{dt} = v\frac{dv}{ds}$.
Substituting $v$,we get $a_t = \left(s\sqrt{\frac{2a}{m}}\right) \frac{d}{ds}\left(s\sqrt{\frac{2a}{m}}\right) = s\left(\frac{2a}{m}\right) = \frac{2as}{m}$.
The net acceleration is $a_{net} = \sqrt{a_R^2 + a_t^2} = \sqrt{\left(\frac{2as^2}{mR}\right)^2 + \left(\frac{2as}{m}\right)^2}$.
Simplifying,$a_{net} = \frac{2as}{m} \sqrt{\frac{s^2}{R^2} + 1}$.
The net force is $F = m \cdot a_{net} = 2as \sqrt{1 + \frac{s^2}{R^2}}$.

(C) In circular motion,the tangential acceleration $\vec a_t$ is directed along the tangent to the path,and the centripetal acceleration $\vec a_c$ is directed towards the center of the circle along the radius.
Since the tangent is perpendicular to the radius,$\vec a_t$ and $\vec a_c$ are perpendicular to each other. Therefore,$\vec a_t \cdot \vec a_c = 0$.
The velocity vector $\vec v$ is always directed along the tangent to the path. Since $\vec a_c$ is directed towards the center (along the radius),$\vec a_c$ is always perpendicular to $\vec v$. Therefore,$\vec a_c \cdot \vec v = 0$.
The tangential acceleration $\vec a_t$ is parallel or anti-parallel to the velocity vector $\vec v$. If the speed is increasing,$\vec a_t$ and $\vec v$ are in the same direction,so $\vec a_t \cdot \vec v > 0$. If the speed is decreasing,$\vec a_t$ and $\vec v$ are in opposite directions,so $\vec a_t \cdot \vec v < 0$. Thus,$\vec a_t \cdot \vec v$ can be positive or negative.
Comparing these with the given options,the statement $\vec a_c \cdot \vec v$ may be positive or negative is incorrect because $\vec a_c \cdot \vec v$ is always zero.

(B) The centripetal acceleration $a_c$ is given by $a_c = v^2/r$.
Given $a_c = 4/r^2$,we equate the two expressions:
$v^2/r = 4/r^2$
$v^2 = 4/r$
$v = 2/\sqrt{r}$
The momentum $P$ of the particle is given by $P = mv$.
Substituting the value of $v$:
$P = m \times (2/\sqrt{r}) = 2m/\sqrt{r}$.

(D) Given the tangential acceleration $a_t = K^2rt^2$.
Since $a_t = \frac{dv}{dt}$,we have $dv = K^2rt^2 dt$.
Integrating both sides,$\int dv = \int K^2rt^2 dt$,which gives $v = \frac{K^2rt^3}{3}$.
Since the velocity $v$ changes with time,the centripetal acceleration $a_c = \frac{v^2}{r}$ is not constant.
The power delivered by a force $\vec{F}$ is given by $P = \vec{F} \cdot \vec{v}$.
The centripetal force $\vec{F}_c$ is always directed towards the center of the circular path,while the velocity vector $\vec{v}$ is always tangential to the path.
Therefore,the angle between $\vec{F}_c$ and $\vec{v}$ is always $90^\circ$.
Thus,the power delivered by the centripetal force is $P_c = \vec{F}_c \cdot \vec{v} = F_c v \cos(90^\circ) = 0$.

(C) The car is undergoing non-uniform circular motion,so it has two components of acceleration: tangential acceleration $(a_t)$ and centripetal (radial) acceleration $(a_c)$.
Given: Speed $v = 40\,m/s$,radius $r = 400\,m$,and tangential acceleration $a_t = 3\,m/s^2$.
The centripetal acceleration is calculated as $a_c = \frac{v^2}{r} = \frac{40^2}{400} = \frac{1600}{400} = 4\,m/s^2$.
The total acceleration $a$ is the vector sum of these perpendicular components: $a = \sqrt{a_c^2 + a_t^2}$.
Substituting the values: $a = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\,m/s^2$.

(D) The centripetal acceleration of a particle moving in a circular path is given by the formula $a_{cp} = \frac{v^2}{r}$.
Initially,the speed is $v$,so the initial centripetal acceleration is $a_1 = \frac{v^2}{r}$.
Finally,the speed is $2v$,so the final centripetal acceleration is $a_2 = \frac{(2v)^2}{r} = \frac{4v^2}{r}$.
Comparing the two,we get $a_2 = 4 \times \left(\frac{v^2}{r}\right) = 4a_1$.
Therefore,the centripetal acceleration has changed by a factor of $4$.

(B) In circular motion,the velocity vector $\vec{v}$ is related to the angular velocity vector $\vec{\omega}$ and the position vector $\vec{r}$ by the relation $\vec{v} = \vec{\omega} \times \vec{r}$.
The acceleration vector $\vec{a}$ is the time derivative of the velocity vector: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(\vec{\omega} \times \vec{r})$.
Using the product rule for differentiation,we get $\vec{a} = \frac{d\vec{\omega}}{dt} \times \vec{r} + \vec{\omega} \times \frac{d\vec{r}}{dt}$.
Since $\frac{d\vec{\omega}}{dt} = \vec{\alpha}$ (angular acceleration) and $\frac{d\vec{r}}{dt} = \vec{v}$,we have $\vec{a} = \vec{\alpha} \times \vec{r} + \vec{\omega} \times \vec{v}$.
For uniform circular motion,the angular acceleration $\vec{\alpha} = 0$,so the acceleration is purely centripetal: $\vec{a}_c = \vec{\omega} \times \vec{v}$.

(D) The angular acceleration is given by $\alpha = \frac{d\omega}{dt} = \frac{d}{dt}(a - bt) = -b$.
The tangential acceleration is $a_t = R\alpha = -Rb$.
At time $t = \frac{2a}{b}$,the angular velocity is $\omega = a - b(\frac{2a}{b}) = a - 2a = -a$.
The radial (centripetal) acceleration is $a_r = R\omega^2 = R(-a)^2 = Ra^2$.
The magnitude of the total acceleration is $a = \sqrt{a_t^2 + a_r^2} = \sqrt{(-Rb)^2 + (Ra^2)^2} = \sqrt{R^2b^2 + R^2a^4} = R\sqrt{a^4 + b^2}$.

(C) The object undergoes non-uniform circular motion,so it has both centripetal acceleration $(a_c)$ and tangential acceleration $(a_t)$.
Given: velocity $v = 30\, m/s$,radius $r = 450\, m$,and tangential acceleration $a_t = 2\, m/s^2$.
The centripetal acceleration is $a_c = \frac{v^2}{r} = \frac{30^2}{450} = \frac{900}{450} = 2\, m/s^2$.
The net acceleration $a$ is given by $a = \sqrt{a_c^2 + a_t^2}$.
Substituting the values: $a = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.828\, m/s^2$.
Thus,the acceleration is nearly $2.8\, m/s^2$.

(A) In non-uniform circular motion,the total acceleration $a$ is the vector sum of the radial acceleration $a_r$ and the tangential acceleration $a_t$. Since these two components are perpendicular to each other,the magnitude of the resultant acceleration is given by $a = \sqrt{a_r^2 + a_t^2}$.
Given $a = 5\, ms^{-2}$.
Checking option $A$: $a_r = 3\, ms^{-2}$ and $a_t = 4\, ms^{-2}$.
$a = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, ms^{-2}$.
This matches the given total acceleration. Thus,option $A$ is correct.

(B) In circular motion,the velocity vector is always directed along the tangent to the circular path,which is also known as the transverse direction.
For a uniformly accelerated circular motion,the particle experiences two types of acceleration:
$1$. Radial (centripetal) acceleration $(a_r = v^2/r)$,which is directed towards the center of the circle.
$2$. Tangential acceleration $(a_t = dv/dt)$,which is directed along the tangent (transverse direction).
Since both radial and tangential accelerations are present,the total acceleration has both radial and transverse components.

(B) The angular momentum is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$. Since the particle is moving in a circular path,the position vector $\vec{r}$ and velocity vector $\vec{v}$ are always perpendicular. The magnitude of angular momentum is $L = mvr$. Since the linear speed $v$ is decreasing,the magnitude of angular momentum $L$ is not conserved.
However,for a particle moving in a circular path in a plane,the direction of the angular momentum vector $\vec{L}$ is perpendicular to the plane of motion (determined by the right-hand rule). As long as the particle remains in the same circular path,the direction of $\vec{L}$ remains constant.
Since the particle is confined to a circular path,it cannot spiral towards the centre. Because the speed is changing,there is a tangential acceleration $a_t$ in addition to the centripetal acceleration $a_c$. Therefore,the net acceleration is not directed towards the centre. Thus,only the direction of angular momentum is conserved.

(A) Length of the string,$r = 80\; cm = 0.8\; m$.
Number of revolutions $= 14$,Time taken $= 25\; s$.
Frequency,$\nu = \frac{\text{Number of revolutions}}{\text{Time taken}} = \frac{14}{25}\; Hz$.
Angular frequency,$\omega = 2\pi\nu = 2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25}\; rad/s = 3.52\; rad/s$.
Centripetal acceleration,$a_c = \omega^2 r = (3.52)^2 \times 0.8$.
$a_c = 12.3904 \times 0.8 = 9.91232\; m/s^2 \approx 9.91\; m/s^2$.
The direction of centripetal acceleration is always directed along the radius towards the center of the circular path.

(N/A) Speed of the cyclist,$v = 27 \; km/h = 7.5 \; m/s$.
Radius of the circular turn,$r = 80 \; m$.
Centripetal acceleration is given as:
$a_c = \frac{v^2}{r} = \frac{(7.5)^2}{80} = 0.703 \; m/s^2 \approx 0.7 \; m/s^2$.
Tangential acceleration is given as $a_T = 0.5 \; m/s^2$.
Since the angle between $a_c$ and $a_T$ is $90^{\circ}$,the resultant acceleration $a$ is given by:
$a = \sqrt{a_c^2 + a_T^2} = \sqrt{(0.7)^2 + (0.5)^2} = \sqrt{0.49 + 0.25} = \sqrt{0.74} \approx 0.86 \; m/s^2$.
Let $\theta$ be the angle of the resultant acceleration with the direction of centripetal acceleration.
$\tan \theta = \frac{a_T}{a_c} = \frac{0.5}{0.7} = 0.714$.
$\theta = \tan^{-1}(0.714) \approx 35.5^{\circ}$ with the direction of centripetal acceleration.

(A) The acceleration of an object moving in a circular path is the centripetal acceleration,given by the formula $a_c = \frac{v^2}{r}$.
Given:
Speed $v = 50\, m s^{-1}$
Radius $r = 250\, m$
Substituting the values into the formula:
$a_c = \frac{(50)^2}{250} = \frac{2500}{250} = 10\, m s^{-2}$.
Thus,the acceleration is $10\, m s^{-2}$.

(A) In a rigid body,all particles rotate about the same axis of rotation with the same angular velocity $\omega$ at any given instant. If the angular velocity $\omega$ of the rigid body is not constant,it means the body has an angular acceleration $\alpha = \frac{d\omega}{dt} \neq 0$. Since the particle is part of a rigid body,it must be rotating about the axis. $A$ particle rotating with a changing angular velocity $\omega$ experiences both centripetal acceleration $(a_c = \omega^2 r)$ and tangential acceleration $(a_t = \alpha r)$. This combination of accelerations characterizes non-uniform circular motion.

(C) In circular motion,the linear acceleration $\vec{a}$ has two components:
$1$. The tangential acceleration $\vec{a}_t$,which is directed along the tangent to the circular path.
$2$. The radial (or centripetal) acceleration $\vec{a}_r$,which is directed towards the center of the circle along the radius.
Since the tangent to a circle at any point is always perpendicular to the radius at that point,the angle between the tangential component $\vec{a}_t$ and the radial component $\vec{a}_r$ is $90^{\circ}$.

(N/A) The radial component of acceleration,also known as centripetal acceleration $(a_r = v^2/r)$,is responsible for changing the direction of the linear velocity of the particle.
The tangential component of acceleration $(a_t = dv/dt)$ is responsible for changing the magnitude (speed) of the linear velocity of the particle.