Question

a wheel on a game show is being pulled band starts to rotate with initial angular speed of 1.5 rad/s . it comes to rest after rotating through 3/4 of a turn . calculate the torque exerted on the wheel if the mass and the radius of the wheel is 6.5 kg and 0.7m respectively .

Answer 1

A torque τ acting through an angle ∆θ does a work W is given by the equation

W = -τ∆θ

The kinemativ equation of angular velocity and angular acceleration is given by the equation

$ω^2=ω_0^2 + 2α∆θ$

The wheel angular acceleration can be found using the kinematics equation as

$α = \frac{ω^2-ω_0^2}{2∆θ}$

$ω_0$ is initial angular speed and ω is final angular speed and ∆θ is angular displacement.

Since wheel come to rest after rotating through ¾ or 0.75 of a turn, its final angular velocity becomes zero (ω = 0).

In each turn wheel rotates 2π rad hence the total change in displacement after come to rest can be expressed as

$∆θ = (0.75)(\frac{2π \;rad}{rev})$

Consider the wheel has a form of disk then moment of inertia of disk is given by the equation

$I = \frac{1}{2}mr^2$

The relation between torque and moment of inertia can be expressed as

$τ = Iα$

Here negative sign indicates torque exerted on the wheel to stop

$τ = (\frac{1}{2}mr^2)(\frac{ω^2-ω_0^2}{2∆θ})$

$τ = -\frac{1}{4}(\frac{mr^2ω_0^2}{∆θ})$

Substitute 6.5 kg for mass of the disk (m), 0.7 m for radius of the disk (r), 1.5 rad/s for $ω_0$ and (0.75)$(\frac{2π \;rad}{rev})$ for ∆θ in the above equation and solve for torque.

$τ = -\frac{1}{4}(\frac{(6.5\;kg)(0.7\;m)^2(1.5 \;rad/s)^2}{0.75(2π \;rad/rev)})$

$τ = -\frac{1}{4} (1.52\;N m)$

$τ = 0.38\;N m$