Question

Q.8
A wheel whose radius is R and moment of
inertia about its own axis is I, rotates freely
about its bwn axis. A rope is wrapped on the
wheel. A body of mass m is suspended from
the free end of the rope. The body is released
from rest. The velocity of the body after falling
a distance h would be -
mah) 1/2
(A) TU
(B) (2mgh) 1/2
(m
+I)
1/2
( 2mgh
? (m+I/r2)
om + 1)2
2mgh​

Answer 1

Answer:

sorry I dont' know this answer

Answer 2

Velocity of body after falling a distance h    $\mathbf{v=\sqrt{\frac{2mgh}{\left (m+ \frac{I}{R^{2}} \right )}}}$

Explanation:

1. Given data

  Mass of object = m (kg)

  Radius of wheel = R (m)

  Mass moment of inertia =  $\boldsymbol{I \ (kg.m^{2})}$

  Gravitational acceleration = $\boldsymbol{g \ (\frac{m}{s^{2}})}$

  Vertical difference between initial and final position = h (m)

  Velocity of object at initial  position = $\boldsymbol{0 \ (\frac{m}{s})}$

  velocity of object at final position = $\boldsymbol{v \ (\frac{m}{s})}$

  Angular velocity of wheel at initial position = $\boldsymbol{0 \ (\frac{rad}{s})}$

  Angular velocity of wheel at final position = $\boldsymbol{\omega \ (\frac{rad}{s})}$

2. Relation between linear velocity and angular velocity is

   $\mathbf{\omega =\frac{v}{R}}$      ...1)

3. Total initial energy of system= Energy of body + Energy of wheel

   $\mathbf{E_{i}=mgh+0=mgh (J)}$      ...2)

4. Total final energy of system= Energy of body + Energy of wheel

   $\mathbf{E_{f}=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega ^{2}}$

   $\mathbf{E_{f}=\frac{1}{2}mv^{2}+\frac{1}{2}I\left ( \frac{v}{R} \right ) ^{2}}$

   $\mathbf{E_{f}=\frac{1}{2}mv^{2}+\frac{1}{2}I\left ( \frac{v^{2}}{R^{2}} \right )}$           ...3)

5. From conservation of energy

   Total final energy = Total initial energy

   $\mathbf{E_{f}=E_{i}}$

   $\mathbf{\frac{1}{2}mv^{2}+\frac{1}{2}I\left ( \frac{v^{2}}{R^{2}} \right )=mgh}$

  $\mathbf{v^{2}\left (m+ \frac{I}{R^{2}} \right )=2mgh}$

  $\mathbf{v^{2}=\frac{2mgh}{\left (m+ \frac{I}{R^{2}} \right )}}$

  So

  $\mathbf{v=\sqrt{\frac{2mgh}{\left (m+ \frac{I}{R^{2}} \right )}}}$    = Velocity of body after falling a distance h