Question

A wheel of mass $10kg$ and radius $20cm$ is rotating at an angular speed of $100$ rev/min when the motor is turned off. Neglecting the friction at the axis, calculate the force that must be applied tangentially to the wheel to bring it to rest in $10$ revolutions. (Consider wheel to be a uniform circular disc)

• A. $0.86N$
• B. $0.87N$
• C. $0.82N$
• D. $0.89N$

Answer 1

The correct option is B $0.87N$

Given: a wheel of mass 10kg and radius 20cm is rotating at an angular speed of 100 rev/min when the motor is turned off.

To find the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions.

Solution:

As per the given criteria,

Rotation of the wheel = 100 revolution / min or = $\frac{5}{3}$ rev/sec

Now ,

Total rev. = 10 revolution

We know,

$0={\left(\frac{10}{6}\right)}^2-2\alpha \times 10⟹\frac{100}{36}=20\alpha ⟹\alpha =\frac{5}{36}rev/se{c}^2$

or $\alpha =2\pi \times \frac{5}{36}⟹\alpha =\frac{10\pi }{30}rad/se{c}^2$

Now, the moment of inertia of the wheel.

$I=\frac{1}{2}m{r}^2⟹I=\frac{1}{2}({0.2}^2\times 10)⟹I=0.2kg{m}^2$

Since the force is F, Now torque

$F=F.r=0.2F$

Now torque $=I\alpha ⟹=0.2\times \frac{10\pi }{36}⟹=\frac{2\pi }{36}$

or $0.2F=\frac{2\pi }{36}⟹F=\frac{10\pi }{36}=0.87N$