Question

1. A uniform disk of mass M = 5.00 kg and radius r = 24.0 cm is mounted on a motor through its center. The motor accelerates the disk uniformly from rest by exerting a constant torque of 1.50 N · m.

(a) What is the time required for the disk to reach an angular speed of 8.50 ✕ 10² rpm?

(b) What is the number of revolutions through which the disk spins before reaching this angular speed?

2. A thin, hollow sphere of mass 1.60 kg and radius 0.560 m is rolling on a horizontal surface with a constant angular speed of 61.0 rpm. Find the total kinetic energy of the sphere.

Answer 1

M = 5 kg ; R = 24 cm ; Torque = 1.5 N m

using , torque = I * ; where I is the moment of inertia ; is the angular acceleration

torque = (1/2) MR^2 *

1.5 = (1/2) * 5 * (0.24)^2 *

= 10.42 rad/sec^2

a) w = 850 rpm

= 850 * 2pi /60

= 89 rad/sec

using ,

w = wo + t

89 = 0 + 10.42*t

t = 8.54 seconds

b) using w^2 = w0^2 + 2

89^2 = 0 + ( 2*10.42* )

= 380.08 rad

= 380.08 / (2pi)

= 60.5 revolutions

2)

total kinetic energy = translational kinetic energy + rotational kinetic energy
k = (1/2)(mv^2) + (1/2)I w^2

= (1/2)m(rw)^2 + (1/2)(2/3)(mr^2)(w^2)

= (1/2)(mr^2w^2) + (1/3)(m)(r^2)(w^2)

= (5/6)mr^2w^2

= (5/6)(1.6)(0.56* 61*2pi/60)^2

= 17.06 Joules

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