A solid disc of mass 48 kg and radius 0.25 m is initially rotating with an angul

Question

A solid disc of mass 48 kg and radius 0.25 m is initially rotating with an angular velocity of 5.6 rev/s. It then undergoes constant negative angular acceleration due to an externally applied torque and comes to rest. This process takes 9.0 seconds. How many revolutions has it made during these 9.0 seconds? What is the moment of inertia of the disc ? Suppose that the externally applied torque in part was provided by a tangential force on the edge of the disc. What is the magnitude of this force?

Explanation / Answer

here,

mass of disc , m = 4.8 kg

radius , r = 0.25 m

initial speed , w0 = 5.6 rev/s = 35.168 rad/s

a)

t = 9 s

let the angular accelration be alpha

w = w0 +alpha * t

0 = 35.168 + alpha * 9

alpha = 3.91 rad/s^2

the angle covered , theta = w0^2 /( 2alpha)

theta = 35.168^2 /( 2 * 3.91)

theta = 158.256 rad

the number of revolutions , n = theta/2pi = 25.2 rev

b)

the moment of inertia , I = 0.5 * m * r^2

I = 0.5 * 4.8 * 0.25^2

I = 0.15 kg.m^2

c)

let the force applied be F

equating the torques

F * r = I * alpha

F * 0.25 = 0.15 * 3.91

F = 2.35 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.