Question

A flywheel of mass 25 kg has a radius of 0.2 m. it is making 240 rpm .what is the torque necessary to bring it to rest in 20 s.if the torque is due to a force applied tangentially, on the rim of wheel, what is the magnitude of the force? assume that mass of flywheel is concentrated at its rim

Answer 1

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Here,

Mass, m = 25 kg

Radius, r = 0.2 m

Initial angular velocity, w1 = 20 rpm = 20 × 2π/60 rad/s = 2π/3 rad/s

Final angular velocity, w2 = 0

Time taken, t = 20 s

Using,

w2 = w1 + αt

=> α = (w2 – w1)/t

=> α = (0 - 2π/3)/20

=> α = -π/30 rad/s2

The negative sign implies angular deceleration.

The moment of inertia of the flywheel is, I = mr2 = (25)(0.22) = 1 kg m2

Now, torque, τ = Iα = (1)(π/30) [considering only magnitude of α]

=> τ = 0.1047 Nm

Now, we know, τ = Fr

=> F = τ/r = 0.1047/0.2 = 0.5235 N

Answer 2

Given :

• Mass = 25 kg
• Radius = 0.2 m
• $\sf n_1 = 240\:rpm= 4\:rps$
• $\sf n_2= 0$
• t = 20 sec

To Find :

• Magnitude of Force

Solution :

Moment of inertia of flywheel

$\tt I = {MR}^{2} \\ \\ \tt I = 25 \times {(0.2)}^{2} \\ \\ \tt I = 1 \: kg \: {m}^{2}$

$\tt \tau = I\alpha \\ \\ \tt \tau = \frac{I(\omega_2 -\omega_1 }{t} \\ \\ \tt \tau = i \frac{2\pi(n_2 - n_1)}{t} \\ \\ \tt \tau = \frac{1 \times 2\pi(0 - 4)}{20} \\ \\ \tt \tau = - 0.4\pi \\ \\ \large\boxed{ \tt \tau = - 1.257 \: \: N -m}$

If F is the magnitude force applied on the rim, then

$\tt F = R \times \tau \\ \\ \tt\therefore F = \frac{ \tau}{r} \\ \\ \tt F= \frac{1.257}{0.2} \\ \\ \large\boxed{ \tt F = 6.285 \: N}$