Question

A grinding wheel is in the form of a uniform solid disk of radius 7.00 cm and mass 2.00 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.600 N,m that the motor exerts on the wheel. (a) How long does the wheel take to reach its final operating speed of 1 200 rev/min? (b) Through how many revolutions does it turn while accelerating?

Answer 1

Answer:

Part a)

$t = 1.03 s$

Part b)

$N = 10.3 rev$

Explanation:

Mass of the disc is given as

m = 2.00 kg

radius of the disc is given as

r = 7 cm

so moment of inertia of the disc is given as

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

$I = \frac{1}{2}(2)(0.07)^2$

$I = 4.9 \times 10^{-3} kg m^2$

Now given that torque on the disc is

$\tau = 0.600 Nm$

so here the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{0.600}{4.9 \times 10^{-3}}$

$\alpha = 122.45 rad/s^2$

Part a)

if disc start from rest and then achieve final speed as 1200 rpm then

$f = 1200/60 = 20 rev/s$

so final speed is

$\omega = 2\pi(20) = 40\pi rad/s$

now the time taken to reach this speed is given as

$\alpha t = \omega$

$(122.45) t = 40\pi$

$t = 1.03 s$

Part b)

Number of revolution in the same time is given as

$N = \frac{\omega_o + \omega}{4\pi} t$

$N = \frac{40\pi + 0}{4\pi}(1.03) = 10.3 rev$