A ball of mass m is attached to the free end | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards. The tension $T$ can be resolved

Vedclass · Accepted Answer

$\sqrt{\frac{T}{m l}}$

Vedclass · Answer

(B) The forces acting on the ball are the tension $T$ in the string and the gravitational force $mg$ acting downwards. The tension $T$ can be resolved into two components: $T \cos 	heta$ acting vertically upwards and $T \sin 	heta$ acting horizontally towards the center of the circular path.For vertical equilibrium:$T \cos 	heta = mg$  --- $(1)$For horizontal circular motion,the centripetal force is provided by the horizontal component of tension:$T \sin 	heta = mr \omega^2$  --- $(2)$From the geometry of the figure,the radius of the circular path is $r = l \sin 	heta$.Substituting $r$ into equation $(2)$:$T \sin 	heta = m(l \sin 	heta) \omega^2$Dividing both sides by $m l \sin 	heta$ (assuming $\sin 	heta 
eq 0$):$\omega^2 = \frac{T}{ml}$Therefore,the angular velocity is:$\omega = \sqrt{\frac{T}{ml}}$

$A$ ball of mass $m$ is attached to the free end of a string of length $l$. The ball is moving in a horizontal circular path about the vertical axis as shown in the diagram. The angular velocity $\omega$ of the ball will be ($T =$ Tension in the string).

Similar Questions

If the string of a conical pendulum makes an angle $\theta$ with the horizontal,then the square of its time period is proportional to

Consider the following statements:
Assertion $(A)$: $A$ cyclist always bends inwards while negotiating a curve.
Reason $(R)$: By bending,he provides the necessary centripetal force by shifting his centre of gravity.

$A$ small disc is placed on the top of a smooth hemisphere of radius $R$. What is the smallest horizontal velocity $v$ that should be given to the disc for it to leave the hemisphere immediately at the top and not slide down it? [There is no friction]

Assume a proton is rotating along a circular path of radius $1 \,m$ under a centrifugal force of $4 \times 10^{-12} \,N$. If the mass of the proton is $1.6 \times 10^{-27} \,kg$, then its angular velocity of rotation is

$A$ car is moving on an overbridge of radius $R$ with a constant speed $v$. As the car is descending on the overbridge from point $B$ to $C$,the normal reaction on it due to the bridge:

Vedclass Test Series

Exam Paper Generator

Online Exam Module