A wheel having a moment of inertia of $5.00 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ constant torque of $3.00 \mathrm {~N} \cdot \mathrm { m }$ 8.00 s?
A) 122 J
B) 57.6 J
C) 78.8 J
D) 64.0 J

Question

Quizplus · Accepted Answer

The work done by a torque is given by the equation $W = \tau \theta$, where $W$ is the work, $\tau$ is the torque, and $\theta$ is the angular displacement in radians. To find the angular displacement, we use the equation $\theta = \omega t + \frac{1}{2} \alpha t^2$, where $\omega$ is the initial angular velocity (which is 0 in this case since it's not mentioned), $\alpha$ is the angular acceleration, and $t$ is the time. The angular acceleration $\alpha$ can be found from the torque and moment of inertia using the equation $\alpha = \frac{\tau}{I}$, where $I$ is the moment of inertia.Given:- $I = 5.00 \, \mathrm{kg \cdot m^2}$- $\tau = 3.00 \, \mathrm{N \cdot m}$- $t = 8.00 \, \mathrm{s}$First, calculate the angular acceleration:$\alpha = \frac{\tau}{I} = \frac{3.00}{5.00} = 0.60 \, \mathrm{rad/s^2}$Since the initial angular velocity $\omega = 0$, the angular displacement $\theta$ over time $t$ is:$\theta = 0 + \frac{1}{2} \alpha t^2 = \frac{1}{2} \cdot 0.60 \cdot (8.00)^2 = 19.2 \, \mathrm{rad}$Finally, calculate the work done:$W = \tau \theta = 3.00 \cdot 19.2 = 57.6 \, \mathrm{J}$

A Wheel Having a Moment of Inertia Of 5.00 kg⋅m25.00 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }5.00 kg⋅m2

A Wheel Having a Moment of Inertia Of $5.00 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$