A uniform rod of mass m and length l is suspe | vedclass

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

Question

(B) Immediately after one string is cut,the rod starts rotating about the point of suspension of the remaining string. The torque $	au$ about this po

Vedclass · Accepted Answer

$mg/4$

Vedclass · Answer

(B) Immediately after one string is cut,the rod starts rotating about the point of suspension of the remaining string. The torque $	au$ about this point is due to the weight $mg$ acting at the center of mass,which is at a distance $l/2$ from the pivot point.$	au = mg \cdot \frac{l}{2}$Using $	au = I \alpha$,where $I = \frac{ml^2}{3}$ is the moment of inertia of the rod about one end:$mg \cdot \frac{l}{2} = \frac{ml^2}{3} \alpha$$\alpha = \frac{3g}{2l}$The linear acceleration of the center of mass $a_c$ is given by $a_c = \frac{l}{2} \alpha = \frac{l}{2} \cdot \frac{3g}{2l} = \frac{3g}{4}$.Applying Newton's second law for the translational motion of the center of mass:$mg - T = m a_c$$T = mg - m \left(\frac{3g}{4}ight)$$T = mg - \frac{3mg}{4} = \frac{mg}{4}$

$A$ uniform rod of mass $m$ and length $l$ is suspended by means of two identical inextensible light strings as shown in the figure. The tension in one string immediately after the other string is cut is . . . . . . . ($g$ is the acceleration due to gravity)

Similar Questions

$A$ cord is wound round the circumference of a wheel of radius $r$. The axis of the wheel is horizontal and the moment of inertia about it is $I$. $A$ weight $mg$ is attached to the end of the cord and falls from rest. After falling through a distance $h$,the angular velocity of the wheel will be

$A$ string is wrapped around a disc of mass $M$ and radius $R$ and the free end is fixed to the ceiling. The centre of mass falls down as the disc unwinds the string. The tension in the string is

What is rotational motion and an axis of rotation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module