Question 46: Centre of Mass and Rotational Motion

Question

A body of mass $$5 \mathrm{~kg}$$ at rest is rotated for $$25 \mathrm{~s}$$ with a constant moment of force $$10 \mathrm{~Nm}$$. Find the work done if the moment of inertia of the body is $$5 \mathrm{~kg} \mathrm{~m}^2$$.

Accepted Answer

To find the work done on the body, we need to understand the relationship between torque, angular acceleration, and work done in rotational motion.
We know the following:
Mass of the body, $$m = 5 \mathrm{~kg}$$
Time of rotation, $$t = 25 \mathrm{~s}$$
Constant moment of force (torque), $$	au = 10 \mathrm{~Nm}$$
Moment of inertia, $$I = 5 \mathrm{~kg} \mathrm{~m}^2$$
First, we find the angular acceleration. Torque ($$	au$$) is related to angular acceleration ($$\alpha$$) by the equation:
$$	au = I \alpha$$
Solving for angular acceleration ($$\alpha$$):
$$\alpha = \frac{	au}{I} = \frac{10 \mathrm{~Nm}}{5 \mathrm{~kg} \mathrm{~m}^2} = 2 \mathrm{~rad/s^2}$$
Next, we need to find the final angular velocity ($$ \omega $$) after time $$t$$. The initial angular velocity ($$ \omega_0 $$) is zero because the body is at rest initially. Using the kinematic equation for rotational motion:
$$ \omega = \omega_0 + \alpha t $$
Substitute the known values:
$$ \omega = 0 + (2 \mathrm{~rad/s^2})(25 \mathrm{~s}) = 50 \mathrm{~rad/s} $$
Now, we can calculate the work done. The work done by a torque in rotational motion is given by:
$$ W = \frac{1}{2} I \omega^2 $$
Substitute the known values:
$$ W = \frac{1}{2} (5 \mathrm{~kg} \mathrm{~m}^2) (50 \mathrm{~rad/s})^2 $$
Calculate the work done:
$$ W = \frac{1}{2} 	imes 5 	imes (50)^2 = \frac{1}{2} 	imes 5 	imes 2500 = 6250 \mathrm{~J} $$
The work done is $$ 6250 \mathrm{~J} $$, which corresponds to Option C.

Answer

625 J

Answer

125 J

Answer

6250 J

Answer

1250 J

Centre of Mass and Rotational Motion

A body of mass $5 \mathrm{~kg} at rest is rotated for 25 \mathrm{~s} with a constant moment of force 10 \mathrm{~Nm}. Find the work done if the moment of inertia of the body is 5 \mathrm{~kg} \mathrm{~m}^2$.

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