Question 112: Centre of Mass and Rotational Motion

Question

A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m , would be:

Accepted Answer

To determine the angular velocity of the wheel after the cord is unwound by 1 meter, we start by calculating the work done by the force applied on the cord.
Work Done by Force (W_F):
$    W_F = 20 \, 	ext{N} 	imes 1 \, 	ext{m} = 20 \, 	ext{J}    $
Relationship with Kinetic Energy:
The work done (W_F) is converted into the change in kinetic energy of the wheel. Therefore, we have:
$    \Delta 	ext{KE} = 20 \, 	ext{J} = \frac{1}{2} I \omega^2    $
where $ I $ is the moment of inertia of the wheel, and $ \omega $ is the angular velocity.
Determine the Moment of Inertia (I):
Since the wheel has mass $ M = 10 \, 	ext{kg} $ and radius $ R = 10 \, 	ext{cm} = 0.1 \, 	ext{m} $, and knowing it's a solid disk:
$    I = MR^2 = 10 	imes 0.1^2 = 0.1 \, 	ext{kg} \cdot 	ext{m}^2    $
Solve for Angular Velocity ($\omega$):
Apply the equation for kinetic energy:
$    20 = \frac{1}{2} 	imes 0.1 	imes \omega^2    $
Solving for $\omega$, we get:
$    20 = 0.05 \omega^2    $
$    \omega^2 = \frac{20}{0.05} = 400    $
$    \omega = \sqrt{400} = 20 \, 	ext{rad/s}    $
Therefore, the angular velocity of the wheel after the cord is unwound by 1 meter is $ 20 \, 	ext{rad/s} $.

Answer

20 rad/s

Answer

30 rad/s

Answer

10 rad/s

Answer

0 rad/s

Centre of Mass and Rotational Motion

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