5. A disk with a mass of 14kg and a diameter of 30cm is mounted on a horizontal

Question

5. A disk with a mass of 14kg and a diameter of 30cm is mounted on a horizontal axle as shown in the figure below. The disk is initially at rest. A constant force F=70N is applied to the edge of the disk at an angle of 37° as shown. After 2s, the force is suddenly reduced to 24N and the disk spins with constant angular velocity. There is friction between the disk and the axle. (a) What is the magnitude of the torque due to friction between the disk and the axle? (b) What is the angular velocity of the disk after 2s? (c) What is the kinetic energy of the disk after 2s? 2

Explanation / Answer

(A) due to force after 2 sec, torque = r x F = r F sin37

= 2.17 N m

now it is moving with constant velocity so nettorque has to be zero.

so magnitude of torque due to friction = 2.17 N m

(B) net torque = (0.15 x 70 xsin37) - 2.17

= 4.15 N m

torque = I alpha

4.15 = (14 x 0.15^2 / 2) alpha

alpha = 26.4 rad/s^2

w= w0 + alpha t

w = 52.8 rad/s

(C) KE = I w^2 /2 = ( M R^2 / 2) w^2 /2

KE = 218.7 J

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

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