Question

A pottery wheel, initially at rest, is assumed to be a uniform disk of mass 3.0 kg and radius 20.0 cm. A 25.0 N force is applied tangentially to the rim.
(a) What is the angular acceleration of this pottery wheel?
(b) If the force is applied for 30.0 sec, find the instantaneous angular velocity at 30.0 sec.
(c) What will be the angular displacement for this wheel from t= 15 s until t= 30s?

Answer 1

Answer:

a) 83.33 rad/s^2

b) 2500 rad/s

c) 6187.5 rad

Explanation:

Given data:

radius of wheel = 20.0 cm

mass of wheel = 3.0 kg

force = 25 N

a) angular acceleration

we know torque is given as

$T = I \times \alpha$

$F\times R = I \times \alpha$

$F\times R = \frac{1}{2} mR^2 \times \alpha$

solving for angular acceleration $\alpha$

$\alpha = \frac{2F}{mR} = \frac{2*25}{3*0.2} = 83.33 rad/s^2$

b)  angular velocity

duartion of force applied = 30.0 sec

$\omega_{30} = \omega_0 +alpha t$

                      = o + 83.33 *30 = 2500 rad/sec

c) angular displacement

$\theta_{15} = = \omega_0 +\frac{1}{2} alpha t_{15}^2 = 0 + \frac{1}{2} 83.33 \times 15^2$

$\theta_{30} = \omega_0 +\frac{1}{2} alpha t_{30}^2 = 0 + \frac{1}{2} 83.33 \times 30^2$

$\Delta \theta = \theta_{30} - \theta_{15}$

$\Delta \theta = 6187.5 rad$