A block of mass m is attached to a pulley dis | vedclass

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

Question

(D) When the string becomes taut,an impulsive tension $T$ acts on both the block and the disc for a very short time interval $\Delta t$.For the block,

Vedclass · Accepted Answer

$10/3\, m/s$

Vedclass · Answer

(D) When the string becomes taut,an impulsive tension $T$ acts on both the block and the disc for a very short time interval $\Delta t$.For the block,the impulse-momentum theorem gives: $-T \Delta t = m v - m v_0$,where $v_0 = 5\, m/s$ is the initial velocity and $v$ is the final velocity.For the disc,the angular impulse-angular momentum theorem about the hinge gives: $(T \cdot r) \Delta t = I \omega$,where $I = \frac{1}{2} m r^2$ is the moment of inertia of the disc.Since the string is taut,the tangential velocity of the disc at the rim must equal the velocity of the block,so $v = \omega r$,or $\omega = v/r$.Substituting $\omega$ into the disc equation: $T \Delta t = \frac{I \omega}{r} = \frac{(\frac{1}{2} m r^2) (v/r)}{r} = \frac{1}{2} m v$.Substituting $T \Delta t$ into the block equation: $-(\frac{1}{2} m v) = m v - m v_0$.Rearranging the terms: $m v_0 = m v + \frac{1}{2} m v = \frac{3}{2} m v$.Solving for $v$: $v = \frac{2}{3} v_0 = \frac{2}{3} 	imes 5 = \frac{10}{3}\, m/s$.

$A$ block of mass $m$ is attached to a pulley disc of equal mass $m$ and radius $r$ by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of $5\, m/s$. Its velocity when the string becomes taut will be

Similar Questions

$A$ uniform rod of mass $m$ and length $l$ hinged at its end is released from rest when it is in the horizontal position. The normal reaction at the hinge when the rod becomes vertical is:

Inner and outer radii of a spool are $r$ and $R$ respectively. $A$ thread is wound over its inner surface and placed over a rough horizontal surface. The thread is pulled by a force $F$ as shown in the figure. Then,in the case of pure rolling:

$A$ rod of mass $m$ and length $l$ is hinged at one end to a horizontal floor and stands vertically. If it is allowed to fall,the velocity with which its upper end strikes the floor is:

The graph between $\log_e L$ and $\log_e P$ will be (where $L$ is angular momentum and $P$ is linear momentum):

Vedclass Test Series

Exam Paper Generator

Online Exam Module