A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed

1.	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» Given in the question :- $\omega$ ROTATION of the wheel = 100 REVOLUTION / MIN or = 5/3 revolution/second. $\omega' = 0$ Now , $\theta$ Total rev. = 10 revolution $\alpha$ Now we have to find = ? 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 :-)*

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»

Given in the question :-

$\omega$ ROTATION of the wheel = 100 REVOLUTION / MIN or = 5/3 revolution/second.

$\omega' = 0$ Now ,

$\theta$ Total rev. = 10 revolution

$\alpha$ Now we have to find = ?

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 :-)