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

(N/A) No,the tangential acceleration is not always zero. The tangential acceleration $a_t$ is defined as the rate of change of the magnitude of velocity,given by $a_t = \frac{dv}{dt} = r \alpha$,where $r$ is the radius and $\alpha$ is the angular acceleration.
If the particle moves with a constant speed (uniform circular motion),then $\frac{dv}{dt} = 0$,which implies $a_t = 0$.
If the particle moves with a variable speed (non-uniform circular motion),then $\alpha \neq 0$,and the tangential acceleration is non-zero.

(A) The acceleration of an object moving in a circular path is given by the centripetal acceleration formula:
$a = \frac{v^2}{r}$
Given:
Speed $v = 4 \, m/s$
Radius $r = 2 \, m$
Substituting the values into the formula:
$a = \frac{(4)^2}{2} = \frac{16}{2} = 8 \, m/s^2$
Thus,the acceleration of the electron is $8 \, m/s^2$.

(A) In circular motion,the acceleration vector $\overrightarrow{a}$ can be resolved into two components: the centripetal acceleration $\overrightarrow{a}_c$ (perpendicular to velocity) and the tangential acceleration $\overrightarrow{a}_t$ (parallel to velocity).
Given $\overrightarrow{v} = 2 \hat{i} \text{ m/s}$ and $\overrightarrow{a} = 2 \hat{i} + 4 \hat{j} \text{ m/s}^2$.
The component of acceleration perpendicular to the velocity is the centripetal acceleration,$a_c$.
Since $\overrightarrow{v}$ is along the $\hat{i}$ direction,the component of $\overrightarrow{a}$ perpendicular to $\overrightarrow{v}$ is the $\hat{j}$ component,which is $a_c = 4 \text{ m/s}^2$.
The formula for centripetal acceleration is $a_c = \frac{v^2}{R}$,where $R$ is the radius of the circle.
Rearranging for $R$,we get $R = \frac{v^2}{a_c}$.
Substituting the values: $R = \frac{(2)^2}{4} = \frac{4}{4} = 1 \text{ m}$.

(A) The power delivered by a force is given by $P = \vec{F} \cdot \vec{v}$.
Since the speed of the particle is increasing,the work done by the net force on the particle must be positive.
In circular motion,the net force $\vec{F}$ can be resolved into two components: the centripetal force $\vec{F}_c$ (directed towards the center) and the tangential force $\vec{F}_t$ (directed along the tangent).
The centripetal force $\vec{F}_c$ is always perpendicular to the velocity $\vec{v}$,so $\vec{F}_c \cdot \vec{v} = 0$.
The tangential force $\vec{F}_t$ is responsible for the change in speed. Since the speed is increasing,$\vec{F}_t$ must be in the same direction as the velocity $\vec{v}$.
Therefore,$\vec{F} \cdot \vec{v} = (\vec{F}_c + \vec{F}_t) \cdot \vec{v} = \vec{F}_c \cdot \vec{v} + \vec{F}_t \cdot \vec{v} = 0 + F_t v = F_t v > 0$.
Thus,$\vec{F} \cdot \vec{v} > 0$.

(C) The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{r} = \frac{20^2}{80} = \frac{400}{80} = 5 \, m/s^2$.
Since the speed is decreasing,the tangential acceleration $a_t$ is directed opposite to the velocity vector,so $a_t = -5 \, m/s^2$.
The total acceleration vector $\vec{a}$ is the vector sum of centripetal acceleration $\vec{a}_c$ (directed towards the center) and tangential acceleration $\vec{a}_t$ (directed opposite to velocity).
The angle $\phi$ between the total acceleration $\vec{a}$ and the tangential acceleration $\vec{a}_t$ is given by $\tan \phi = \frac{a_c}{|a_t|} = \frac{5}{5} = 1$,which means $\phi = 45^{\circ}$.
Since the tangential acceleration $\vec{a}_t$ is directed opposite to the velocity $\vec{v}$,the angle between $\vec{a}_t$ and $\vec{v}$ is $180^{\circ}$.
The angle $\theta$ between the total acceleration $\vec{a}$ and the velocity $\vec{v}$ is $\theta = 180^{\circ} - 45^{\circ} = 135^{\circ}$.

(B) Given,$s = 2t^3$ and radius $R = 12 \, m$.
The velocity $v$ is the rate of change of distance along the path: $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3) = 6t^2$.
At $t = 2 \, s$,the velocity $v = 6(2)^2 = 24 \, m/s$.
The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{R} = \frac{24^2}{12} = \frac{576}{12} = 48 \, m/s^2$.
The tangential acceleration $a_t$ is the rate of change of speed: $a_t = \frac{dv}{dt} = \frac{d}{dt}(6t^2) = 12t$.
At $t = 2 \, s$,the tangential acceleration $a_t = 12(2) = 24 \, m/s^2$.
The ratio of tangential to centripetal acceleration is $\frac{a_t}{a_c} = \frac{24}{48} = \frac{1}{2}$,which is $1: 2$.

(C) In non-uniform circular motion,the total acceleration $\vec{a}$ is the vector sum of the centripetal acceleration $\vec{a}_c$ and the tangential acceleration $\vec{a}_T$,i.e.,$\vec{a} = \vec{a}_c + \vec{a}_T$.
$1$. The centripetal acceleration $\vec{a}_c$ is always directed towards the center of the circle,making an angle of $90^{\circ}$ with the velocity vector $\vec{v}$.
$2$. The tangential acceleration $\vec{a}_T$ acts along the tangent. Since the speed is decreasing,$\vec{a}_T$ is directed opposite to the velocity vector $\vec{v}$,making an angle of $180^{\circ}$ with it.
$3$. The resultant acceleration $\vec{a}$ lies between the directions of $\vec{a}_c$ and $\vec{a}_T$. Therefore,the angle $\theta$ between the velocity $\vec{v}$ and the total acceleration $\vec{a}$ must be greater than $90^{\circ}$ (due to $\vec{a}_c$) and less than $180^{\circ}$ (due to $\vec{a}_T$).
Thus,$90^{\circ} < \theta < 180^{\circ}$.

(B) Given: Radius $R = 600\,m$,initial speed $u = 54\,km/h = 15\,m/s$.
Tangential acceleration $a_t = \frac{dv}{dt} = v\frac{dv}{ds}$.
Centripetal acceleration $a_c = \frac{v^2}{R}$.
Given $a_t = a_c$,so $v\frac{dv}{ds} = \frac{v^2}{R} \Rightarrow \frac{dv}{v} = \frac{ds}{R}$.
Integrating both sides: $\int_{15}^{v} \frac{dv}{v} = \int_{0}^{s} \frac{ds}{R} \Rightarrow \ln(\frac{v}{15}) = \frac{s}{R} \Rightarrow v = 15e^{s/R}$.
Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = 15e^{s/R} \Rightarrow e^{-s/R} ds = 15 dt$.
To find the time $T$ for the first quarter revolution,$s$ goes from $0$ to $\frac{\pi R}{2}$.
$\int_{0}^{\pi R/2} e^{-s/R} ds = \int_{0}^{T} 15 dt$.
$[-R e^{-s/R}]_{0}^{\pi R/2} = 15T$.
$-R(e^{-\pi/2} - 1) = 15T \Rightarrow R(1 - e^{-\pi/2}) = 15T$.
$T = \frac{600}{15}(1 - e^{-\pi/2}) = 40(1 - e^{-\pi/2})$.
Comparing with $t(1 - e^{-\pi/2})$,we get $t = 40$.

(C) Given: Radius $R = 180 \, cm = 1.8 \, m$. Frequency $f = \frac{28}{60} \, Hz$. Angular velocity $\omega = 2 \pi f = 2 \times \frac{22}{7} \times \frac{28}{60} = \frac{44}{15} \, rad/s$. The centripetal acceleration is $a = \omega^2 R$. Substituting the values: $a = \left(\frac{44}{15}\right)^2 \times 1.8 = \frac{1936}{225} \times 1.8$. Simplifying: $a = \frac{1936 \times 1.8}{225} = \frac{1936}{125} \, m s^{-2}$. Comparing this with $\frac{1936}{x} \, m s^{-2}$,we get $x = 125$.

(A) Given that the masses of the two particles are equal,so $m_1 = m_2 = m$.
The ratio of their radii of curvature is given as $\frac{r_1}{r_2} = \frac{3}{4}$.
The centripetal force $F$ acting on a particle is given by the formula $F = \frac{mv^2}{r}$.
Since the centripetal force is constant for both particles,we have $F_1 = F_2$.
Substituting the formula,we get $\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$.
Since $m_1 = m_2$,the equation simplifies to $\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2}$.
Rearranging the terms to find the ratio of velocities,we get $\frac{v_1^2}{v_2^2} = \frac{r_1}{r_2}$.
Taking the square root on both sides,we get $\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}}$.
Substituting the given ratio $\frac{r_1}{r_2} = \frac{3}{4}$,we get $\frac{v_1}{v_2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Therefore,the ratio of their velocities is $\sqrt{3}:2$.

(A) Given that the normal acceleration $a_c = \frac{v^2}{r}$ and tangential acceleration $a_t = \frac{dv}{dt}$ are equal:
$\frac{v^2}{r} = \frac{dv}{dt}$
Integrating with respect to time from $t=0$ $(v=4 \ m/s)$ to $t$:
$\int_{4}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{r}$
$\left[ -\frac{1}{v} \right]_{4}^{v} = \frac{t}{r}$
$-\frac{1}{v} + \frac{1}{4} = \frac{t}{0.5} = 2t$
$\frac{1}{v} = \frac{1}{4} - 2t = \frac{1-8t}{4} \implies v = \frac{4}{1-8t}$
Since $v = \frac{ds}{dt}$,we integrate to find the distance $s$ for one revolution $(s = 2\pi r = 2\pi(0.5) = \pi \ m)$:
$\int_{0}^{\pi} ds = \int_{0}^{t} \frac{4}{1-8t} dt$
$\pi = 4 \left[ \frac{\ln(1-8t)}{-8} \right]_{0}^{t}$
$\pi = -\frac{1}{2} \ln(1-8t)$
$-2\pi = \ln(1-8t)$
$e^{-2\pi} = 1-8t$
$8t = 1 - e^{-2\pi}$
$t = \frac{1}{8} [1 - e^{-2\pi}] \ s$
Comparing with the given form,$\alpha = 8$.

(B) Given: Initial velocity $u = 0$,tangential acceleration $a_t = 12 \ m/s^2$,and radius $r = 3 \ m$.
The total acceleration vector $\vec{a}$ is the vector sum of radial acceleration $\vec{a}_c$ and tangential acceleration $\vec{a}_t$.
The angle $\theta$ between the total acceleration and the radial acceleration is given by $\tan \theta = \frac{a_t}{a_c}$.
Given $\theta = 45^{\circ}$,so $\tan 45^{\circ} = 1$,which implies $a_t = a_c$.
We know $a_c = \frac{v^2}{r}$,where $v$ is the instantaneous velocity.
Since $a_t = 12 \ m/s^2$ and $a_c = \frac{v^2}{3}$,we have $12 = \frac{v^2}{3} \Rightarrow v^2 = 36 \Rightarrow v = 6 \ m/s$.
Using the equation of motion $v = u + a_t t$,we get $6 = 0 + 12 \times t$.
Therefore,$t = \frac{6}{12} = 0.5 \ s$ or $(1/2) \ s$.

(A) The total acceleration vector makes an angle of $135^{\circ}$ with the positive $x$-axis (tangent direction). The centripetal acceleration $a_c$ is directed towards the center (negative $x$-direction).
The component of the total acceleration $a = 8\sqrt{2} \ m/s^2$ along the radial direction is $a_c = a \cos(180^{\circ} - 135^{\circ}) = 8\sqrt{2} \cos(45^{\circ}) = 8\sqrt{2} \times \frac{1}{\sqrt{2}} = 8 \ m/s^2$.
We know that the centripetal acceleration is given by $a_c = \frac{v^2}{r}$.
Substituting the values,$8 = \frac{v^2}{2}$.
$v^2 = 16$.
$v = 4 \ m/s$.

(C) Tangential acceleration is given by $a_t = \alpha r$.
Radial (centripetal) acceleration is given by $a_r = \frac{V^2}{r}$.
The ratio of tangential acceleration to radial acceleration is $\frac{a_t}{a_r} = \frac{\alpha r}{V^2 / r}$.
Simplifying this expression,we get $\frac{a_t}{a_r} = \frac{\alpha r^2}{V^2}$.

(C) The tangential acceleration $(a_t)$ is given by the formula $a_t = \alpha r$.
The radial (centripetal) acceleration $(a_r)$ is given by the formula $a_r = \frac{u^2}{r}$.
The ratio of tangential acceleration to radial acceleration is $\frac{a_t}{a_r} = \frac{\alpha r}{u^2 / r}$.
Simplifying this expression,we get $\frac{\alpha r^2}{u^2}$.

(A) In non-uniform circular motion,the tangential acceleration is given by $a_t = r \alpha$.
The radial (centripetal) acceleration is given by $a_r = \frac{v^2}{r}$.
The ratio of tangential to radial acceleration is $\frac{a_t}{a_r} = \frac{r \alpha}{v^2 / r}$.
Simplifying this expression,we get $\frac{a_t}{a_r} = \frac{r^2 \alpha}{v^2}$.

(B) Given: Angular acceleration $\alpha = 4 \ rad/s^2$. Initial angular velocity $\omega_0 = 0$.
$1$. The angular velocity at time $t$ is given by $\omega = \omega_0 + \alpha t = \alpha t$.
$2$. The centrifugal (radial) acceleration is $a_c = \omega^2 r = (\alpha t)^2 r = \alpha^2 t^2 r$.
$3$. The tangential acceleration is $a_t = \alpha r$.
$4$. We are given that the magnitudes are equal: $a_c = a_t$.
$5$. Substituting the expressions: $\alpha^2 t^2 r = \alpha r$.
$6$. Dividing both sides by $\alpha r$ (assuming $\alpha, r \neq 0$): $\alpha t^2 = 1$.
$7$. Solving for $t$: $t^2 = \frac{1}{\alpha} = \frac{1}{4}$.
$8$. Therefore,$t = \sqrt{\frac{1}{4}} = 0.5 \ s$.

(B) Given: Radius $r = 100 \ m$,velocity $v = 20 \ m/s$,and tangential acceleration $a_{t} = 3 \ m/s^{2}$.
First,calculate the radial (centripetal) acceleration $a_{r}$ using the formula $a_{r} = \frac{v^{2}}{r}$.
$a_{r} = \frac{(20)^{2}}{100} = \frac{400}{100} = 4 \ m/s^{2}$.
The resultant acceleration $a$ is the vector sum of the tangential and radial accelerations,which are perpendicular to each other.
$a = \sqrt{a_{r}^{2} + a_{t}^{2}}$.
$a = \sqrt{(4)^{2} + (3)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 \ m/s^{2}$.
Therefore,the resultant acceleration is $5 \ m/s^{2}$.

(B) In non-uniform circular motion, the car possesses two components of acceleration:
$1$. Tangential acceleration $(a_t)$: Given as $2 \,m/s^2$.
$2$. Centripetal (radial) acceleration $(a_c)$: Calculated as $a_c = \frac{v^2}{r} = \frac{(30)^2}{500} = \frac{900}{500} = 1.8 \,m/s^2$.
Since these two components are perpendicular to each other, the net acceleration $(a_{net})$ is given by:
$a_{net} = \sqrt{a_t^2 + a_c^2}$
$a_{net} = \sqrt{(2)^2 + (1.8)^2}$
$a_{net} = \sqrt{4 + 3.24} = \sqrt{7.24} \approx 2.69 \,m/s^2 \approx 2.7 \,m/s^2$.

(C) The equation of the trajectory is $(x-2)^2 + y^2 = 25$. This represents a circle with center at $(2, 0)$ and radius $R = 5 \text{ m}$.
In uniform circular motion,the acceleration is purely centripetal,directed towards the center of the circle.
The magnitude of centripetal acceleration is $a_c = \frac{v^2}{R}$.
Given $v = 2 \text{ m/s}$ and $R = 5 \text{ m}$,we have $a_c = \frac{2^2}{5} = \frac{4}{5} = 0.8 \text{ m/s}^2$.
The particle attains the lowest $y$ coordinate at the point $(2, -5)$.
At this point,the center of the circle $(2, 0)$ is directly above the particle (along the positive $y$-axis).
Therefore,the centripetal acceleration vector is directed along the positive $y$-axis,which is $0.8 \hat{j} \text{ m/s}^2$.

(D) Given: Speed $v = 30 \,ms^{-1}$, Radius $r = 500 \,m$, Tangential acceleration $a_T = 2 \,ms^{-2}$.
The total acceleration $a$ of a particle in circular motion is the vector sum of tangential acceleration $a_T$ and centripetal (radial) acceleration $a_r$.
First, calculate the centripetal acceleration $a_r = \frac{v^2}{r} = \frac{30^2}{500} = \frac{900}{500} = 1.8 \,ms^{-2}$.
The total acceleration is given by $a = \sqrt{a_T^2 + a_r^2}$.
Substituting the values: $a = \sqrt{2^2 + 1.8^2} = \sqrt{4 + 3.24} = \sqrt{7.24}$.
Calculating the square root: $a \approx 2.69 \,ms^{-2}$, which is approximately $2.7 \,ms^{-2}$.

(D) Given: Velocity of the car, $v = 30 \,ms^{-1}$. Radius of the circular road, $r = 300 \,m$. Tangential acceleration, $a_t = 4 \,ms^{-2}$.
First, calculate the centripetal acceleration $(a_c)$ using the formula $a_c = \frac{v^2}{r}$.
$a_c = \frac{30^2}{300} = \frac{900}{300} = 3 \,ms^{-2}$.
The net acceleration $(a)$ is the vector sum of the tangential acceleration and the centripetal acceleration, which are perpendicular to each other.
$a = \sqrt{a_t^2 + a_c^2}$.
$a = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,ms^{-2}$.

(A) Given: Radius, $r = 10 \,m$.
Centripetal acceleration, $a_c = 10 \,ms^{-2}$.
We know that the formula for centripetal acceleration is $a_c = \omega^2 r$.
Rearranging the formula to solve for angular speed $\omega$:
$\omega = \sqrt{\frac{a_c}{r}}$
Substituting the given values:
$\omega = \sqrt{\frac{10 \,ms^{-2}}{10 \,m}} = \sqrt{1 \,s^{-2}} = 1 \,rad \,s^{-1}$.
Therefore, the angular speed is $1 \,rad \,s^{-1}$.

(D) Given: Mass $m = 5 \ kg$,Radius $r = 1 \ m$,Angular velocity $\omega = 2 \ rad \ s^{-1}$.
Centripetal force $F_c$ is given by the formula $F_c = m \omega^2 r$.
Substituting the given values:
$F_c = 5 \times (2)^2 \times 1$
$F_c = 5 \times 4 \times 1$
$F_c = 20 \ N$.
Therefore,the centripetal force is $20 \ N$.

(A) Given,velocity of particle is $\vec{v} = x \hat{i} + yt \hat{j}$.
The magnitude of velocity is $v = \sqrt{x^2 + y^2 t^2}$.
The tangential acceleration $a_t$ is the rate of change of speed:
$a_t = \frac{dv}{dt} = \frac{1}{2\sqrt{x^2 + y^2 t^2}} \cdot (2y^2 t) = \frac{y^2 t}{\sqrt{x^2 + y^2 t^2}}$.
Substituting $t = \frac{x \sqrt{3}}{y}$:
$a_t = \frac{y^2 (x \sqrt{3} / y)}{\sqrt{x^2 + y^2 (3x^2 / y^2)}} = \frac{xy \sqrt{3}}{\sqrt{x^2 + 3x^2}} = \frac{xy \sqrt{3}}{2x} = \frac{\sqrt{3} y}{2}$.
The total acceleration vector is $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(x \hat{i} + yt \hat{j}) = y \hat{j}$.
The magnitude of total acceleration is $a = |\vec{a}| = y$.
The normal acceleration $a_n$ is given by $a_n = \sqrt{a^2 - a_t^2}$.
$a_n = \sqrt{y^2 - (\frac{\sqrt{3} y}{2})^2} = \sqrt{y^2 - \frac{3y^2}{4}} = \sqrt{\frac{y^2}{4}} = \frac{y}{2}$.
Thus,the tangential and normal accelerations are $\frac{\sqrt{3} y}{2}$ and $\frac{y}{2}$ respectively.

(D) The radial (centripetal) acceleration is given by $a_r = \frac{v^2}{R}$.
Given $v = K\sqrt{s}$,we have $a_r = \frac{(K\sqrt{s})^2}{R} = \frac{K^2 s}{R}$.
The tangential acceleration is $a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$.
Since $v = K s^{1/2}$,then $\frac{dv}{ds} = K \cdot \frac{1}{2} s^{-1/2} = \frac{K}{2\sqrt{s}}$.
Thus,$a_t = (K\sqrt{s}) \cdot \left( \frac{K}{2\sqrt{s}} \right) = \frac{K^2}{2}$.
The angle $\theta$ between the total acceleration and the tangential acceleration is given by $\tan \theta = \frac{a_r}{a_t}$.
Substituting the values,$\tan \theta = \frac{K^2 s / R}{K^2 / 2} = \frac{2s}{R}$.

(A) The centripetal acceleration $a_c$ of a particle moving in a circular path of radius $R$ with angular speed $\omega$ is given by $a_c = R \omega^2$.
Let the initial angular speed be $\omega_1 = \omega$ and the final angular speed be $\omega_2 = 2\omega$.
The initial centripetal acceleration is $a_1 = R \omega^2$.
The final centripetal acceleration is $a_2 = R \omega_2^2 = R (2\omega)^2 = 4 R \omega^2$.
Comparing the two,we get $a_2 = 4 a_1$.
Therefore,the centripetal acceleration becomes $4$ times the initial value.

(B) The car experiences two types of acceleration while moving on a circular path with increasing speed:
$1$. Radial (centripetal) acceleration: $a_r = \frac{V^2}{r}$,which is directed towards the center of the circular path.
$2$. Tangential acceleration: $a_t = a$,which is directed along the tangent to the path.
Since these two accelerations are perpendicular to each other,the resultant acceleration $a_R$ is given by the vector sum:
$a_R = \sqrt{a_r^2 + a_t^2}$
Substituting the values:
$a_R = \sqrt{(\frac{V^2}{r})^2 + a^2}$
$a_R = \sqrt{\frac{V^4}{r^2} + a^2}$

(A) Given,$\theta = a t^2$.
The angular velocity is $\omega = \frac{d\theta}{dt} = 2at$.
The tangential velocity is $v = \omega r = 2atr$. Thus,$\frac{v}{t} = 2ar$.
The tangential acceleration is $a_t = \frac{dv}{dt} = 2ar$.
The radial (centripetal) acceleration is $a_n = \frac{v^2}{r} = \frac{(2atr)^2}{r} = 4a^2t^2r$.
The total acceleration is $a_{\text{total}} = \sqrt{a_t^2 + a_n^2} = \sqrt{(2ar)^2 + (4a^2t^2r)^2}$.
$a_{\text{total}} = \sqrt{4a^2r^2 + 16a^4t^4r^2} = 2ar \sqrt{1 + 4a^2t^4}$.
Since $2ar = \frac{v}{t}$,we have $a_{\text{total}} = \frac{v}{t} \sqrt{1 + 4a^2t^4}$.

(B) Given speed $v(t) = at$,where $a = 2 \ m/s^2$.
In circular motion,$v = r \omega$,so $\omega = \frac{v}{r} = \frac{at}{r}$.
Since $\omega = \frac{d\theta}{dt}$,we have $d\theta = \frac{at}{r} dt$.
Integrating for $n$ rounds,$\theta = 2\pi n = \int_0^t \frac{at}{r} dt = \frac{at^2}{2r}$.
Thus,$t^2 = \frac{4\pi nr}{a}$.
Radial acceleration $a_r = \frac{v^2}{r} = \frac{(at)^2}{r} = \frac{a^2 t^2}{r} = \frac{a^2}{r} \cdot \frac{4\pi nr}{a} = 4\pi na$.
Tangential acceleration $a_t = \frac{dv}{dt} = a$.
Total acceleration $A = \sqrt{a_t^2 + a_r^2} = \sqrt{a^2 + (4\pi na)^2} = a \sqrt{1 + (4\pi n)^2}$.
Given $a = 2$ and $n = 2$,$A = 2 \sqrt{1 + (4 \cdot \pi \cdot 2)^2} = 2 \sqrt{1 + 64\pi^2}$.

(D) Initial speed $V = 36 \,km/h = 36 \times \frac{5}{18} = 10 \,m/s$.
Radial acceleration $a_r = \frac{V^2}{R} = \frac{10^2}{50} = \frac{100}{50} = 2 \,m/s^2$.
Tangential acceleration $a_t = 0.5 \,m/s^2$.
Net acceleration $a_{net} = \sqrt{a_r^2 + a_t^2} = \sqrt{2^2 + 0.5^2} = \sqrt{4 + 0.25} = \sqrt{4.25} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \,m/s^2$.
Direction $\theta$ with respect to the radial vector is given by $\tan \theta = \frac{a_t}{a_r} = \frac{0.5}{2} = \frac{1}{4}$.
However, the angle with respect to the tangential vector is $\tan \phi = \frac{a_r}{a_t} = \frac{2}{0.5} = 4$, so $\phi = \tan^{-1}(4)$.

(D) The body experiences two components of acceleration: centripetal acceleration $(a_c)$ and tangential acceleration $(a_T)$.
Centripetal acceleration is given by $a_c = \frac{v^2}{r} = \frac{(2\sqrt{5})^2}{5} = \frac{20}{5} = 4 \,m/s^2$.
Tangential acceleration is given as $a_T = 3 \,m/s^2$.
The net acceleration is $a = \sqrt{a_c^2 + a_T^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,m/s^2$.
The magnitude of the net force is $F = m \times a = 2 \,kg \times 5 \,m/s^2 = 10 \,N$.