Question

A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed of 100 revolution/min when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions ?

Answer 1

Given in the question :-

Rotation of the wheel = 100 revolution / min or $\omega$= 5/3 revolution/second.

Now , $\omega' = 0$

Total rev. $\theta$ = 10 revolution

Now we have to find $\alpha$ = ?

We know the formula,

$\omega' =\omega ^2 - 2 \alpha \theta$

0 = (10/6)² - 2 × α × 10

0 = 100/36 - 20α

α = 5/36 rev/second²

or α = 2 π × (5/36)

α = 10π / 36 radian/second² .

Now, we have to moment of inertia of the wheel.

$I = mr^2 /2$

I = 1/2 (0.2² × 10)

I = 0.2 kg.m²

Since the force is F, Now torque

F = F.r

0.2 F

Now torque = I α

0.20 / (10 π /36)

= 2 π /36

0.2 F = 2 π /36

F = 2 π/7.2

F = 0.87 N

Hope it Helps :-)

Answer 2

The force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions is 8.7N.    

Explanation:

Step 1:

Given values in the question :

Rotation of the wheel = 100 revolution / min

                                              (or)

Rotation of the wheel = 5/3 revolution/second.

Now,  

            w’=0

            Total rev. θ = 10 revolution

Now we have to find = ?

Step 2:

We know the formula,

                $\omega^{\prime}=\omega^{2}-2 \alpha \theta$

                $0=\left(\frac{10}{6}\right)^{2}-2 \times \alpha \times 10$

                $0=\frac{100}{36}-20 \alpha$

                $2 a=\frac{100}{36}$

                 $\alpha=\frac{100}{36 \times 2}$    

                 $\alpha=\frac{50}{26} \mathrm{rev} / \mathrm{second}^{2}$

                          (or)

                $\alpha=2 \pi \times\left(\frac{50}{36}\right)$

                $\alpha=\frac{100 \pi}{36} \text { radian/second }^{2}$

Step 3:

Now, we've got to the wheel moment of inertia.

                     $\begin{aligned}&I=\frac{m r^{2}}{2}\\&I=\frac{1}{2}\left(0.2^{2} \times 10\right)\\&\mathrm{I}=\frac{1}{2}(0.04 \times 10)\\&\mathrm{I}=\frac{1}{2}(0.04)\end{aligned}$

                    $\mathrm{I}=0.2 \mathrm{kg} . \mathrm{m}^{2}$

Step 4:

Because the force is F, the torque now

                   F = F.r

                     = 0.2 F

Now, torque = I α

                   $\begin{aligned}&=0.20 \frac{100 \pi}{36}\\&=\frac{20 \pi}{36}\end{aligned}$

           $0.2 \mathrm{F}=\frac{20 \pi}{36}$

               $F=\frac{20 \pi}{36 \times 0.2}$

               $F=\frac{20 \times 3.14}{36 \times 0.2}$

               $F=\frac{62.8}{7.2}$

         ∴,   F = 8.7 N

Hence, the force that must be applied tangentially to the wheel is,

F = 8.7N.