A uniform disk of mass M = 5.2 kg has a radius of 0.11 m and is pivoted so that

Question

A uniform disk of mass M = 5.2 kg has a radius of 0.11 m and is pivoted so that it rotates freely about its axis. A string wrapped around the disk is pulled with a force F equal to 22 N.

(a) What is the torque being exerted by this force about the rotation axis?
N ? m

(b) What is the angular acceleration of the disk?
rad/s2

(c) If the disk starts from rest, what is its angular speed after 5.3 s?
rad/s

(d) What is its kinetic energy after the 5.3 s?
J

(e) What is the angular displacement of the disk during the 5.3 s?
rad

A uniform disk of mass M = 5.2 kg has a radius of 0.11 m and is pivoted so that it rotates freely about its axis. A string wrapped around the disk is pulled with a force F equal to 22 N. (a) What is the torque being exerted by this force about the rotation axis? (b) What is the angular acceleration of the disk? (c) If the disk starts from rest, what is its angular speed after 5.3 s? (d) What is its kinetic energy after the 5.3 s? (e) What is the angular displacement of the disk during the 5.3 s?

Explanation / Answer

Given data:

the mass of the disk, M = 5.2 kg

radius of the disk , R = 0.11 m

the force F = 22 N

(a)

T(torque) = F R = 22 * 0.11 = 2.42 N-m

the torque being exerted by this force about the rotation axis is 2.42 N-m.

(b)

the moment of inertia of the disk,

I = 1/2 M R2 = 1/2*5.2 *0.112 = 0.03146 kgm2

the angular acceleration,

? = T / I = 2.42 /0.03146 = 76.92 rad/ s2

the angular acceleration of the disk is 96.92 rad/s^2

(c) the angular speed of the disk after 5.3 s,

? = ? t = 76.92*5.3 = 407.676 rad/ s

(d)

the kinetic energy,

KE = 1/2 I ?2 = (0.03146/ 2)* 407.6762 = 2614.32 J

(e)

the angular displacement,

? = 1/2 ? t2 = (76.92 / 2) * 5.3^2 = 1080.34 rad

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