A uniform disk of mass M = 4.6 kg has a radius of 0.11 m and is pivoted so that

Question

A uniform disk of mass M = 4.6 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 23 N.
(a) What is the torque being exerted by this force about the rotation axis?
mark N · m

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

(c) If the disk starts from rest, what is its angular speed after 4.6 s?
mark rad/s

(d) What is its kinetic energy after the 4.6 s?
mark J

(e) What is the angular displacement of the disk during the 4.6 s?
mark rad

Explanation / Answer

Given data: the mass of the disk, M = 4.6 kg radius of the disk , R = 0.11 m the force F = 23 N (a) T(torque) = F R = 23 * 0.11 = 2.53 N-m the torque being exerted by this force about the rotation axis is 2.53 N-m. (b) the moment of inertia of the disk, I = .5 M R2 = (1/2)*(4.6) *(0.112) = 0.02783 kgm2 the angular acceleration, = T / I = 2.53 /0.02783 = 90.90 rad/ s2 the angular acceleration of the disk is 90.90 rad/s^2 (c) the angular speed of the disk after 4.6 s, = t = 90.90*4.6 = 418.14 rad/ s (d) the kinetic energy, KE = (1/2) I 2 = (0.02783/ 2)* 418.142 = 2432.91 J (e) the angular displacement, = 1/2 t2 = (90.90 / 2) * 4.6^2 = 961.722 rad Hope that helps, let me know if you have any questions. Good luck! Given data: the mass of the disk, M = 4.6 kg radius of the disk , R = 0.11 m the force F = 23 N (a) T(torque) = F R = 23 * 0.11 = 2.53 N-m the torque being exerted by this force about the rotation axis is 2.53 N-m. (b) the moment of inertia of the disk, I = .5 M R2 = (1/2)*(4.6) *(0.112) = 0.02783 kgm2 the angular acceleration, the angular acceleration, = T / I = 2.53 /0.02783 = 90.90 rad/ s2 the angular acceleration of the disk is 90.90 rad/s^2 (c) the angular speed of the disk after 4.6 s, the angular acceleration of the disk is 90.90 rad/s^2 (c) the angular speed of the disk after 4.6 s, = t = 90.90*4.6 = 418.14 rad/ s (d) the kinetic energy, KE = (1/2) I 2 = (0.02783/ 2)* 418.142 = 2432.91 J (e) the angular displacement, = 1/2 t2 = (90.90 / 2) * 4.6^2 = 961.722 rad Hope that helps, let me know if you have any questions. Good luck!

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