Question 396: Centre of Mass and Rotational Motion

Question

A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of $2 \mathrm{rad} / \mathrm{s}^2$. Moment of intertia of the wheel is___________ $\mathrm{kg} \mathrm{}\,\, \mathrm{m}^2$. (Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Step 1: Use the Torque Formula
The force pulls the rim and creates a turning effect called torque. The formula for torque is: torque = force ÃÂ radius (F ÃÂ R).
Step 2: Relate Torque to Angular Acceleration
Torque is also described as: torque = moment of inertia ÃÂ angular acceleration (I ÃÂ ÃÂ±).
Step 3: Set Up the Equation
Set the two expressions for torque equal. So, $ F 	imes R = I 	imes \alpha $.
Step 4: Solve for Moment of Inertia $ I $
$ I = \frac{F 	imes R}{\alpha} $
Step 5: Substitute the Values
Here, the force $ F = 10\,\mathrm{N} $, the radius $ R = 0.2\,\mathrm{m} $, and the angular acceleration $ \alpha = 2\,\mathrm{rad/s}^2 $.
Step 6: Calculate
$ I = \frac{10 	imes 0.2}{2} = \frac{2}{2} = 1\,\mathrm{kg\,m}^2 $
So, the moment of inertia of the wheel is $ 1\,\mathrm{kg\,m}^2 $.

Centre of Mass and Rotational Motion

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