r = 0.35 mJ = 6 kg.m?30°F = 12 NFigure 2(a) A force of 12 N is applied to a cylinder of radius r = 0.35 m, as shown in Figure Q2(a). Themoment of inertia of this cylinder is J = 6 kg.m². Determine:Q2 (a)(i)The magnitude and direction of the torque produced on the cylinder.(ii) The angular acceleration a of the cylinder.

Answered: Image /qna-images/answer/10875d0a-26ff-4ecb-8ce4-8ed9ead64065.jpg

A force of 12 N is applied to a cylinder of radius r = 0.35 m, as shown in Figure Q2(a). The moment of inertia of this cylinder is J = 6 kg.m². Determine: The magnitude and direction of the torque produced on the cylinder. Q2 (a) (i) (ii) The angular acceleration a of the cylinder.

A force of 12 N is applied to a cylinder of radius r = 0.35 m, as shown in Figure Q2(a). The moment of inertia of this cylinder is J = 6 kg.m². Determine: The magnitude and direction of the torque produced on the cylinder. Q2 (a) (i) (ii) The angular acceleration a of the cylinder.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Question

r = 0.35 m
J = 6 kg.m?
30°
F = 12 N
Figure 2(a)

A force of 12 N is applied to a cylinder of radius r = 0.35 m, as shown in Figure Q2(a). The
moment of inertia of this cylinder is J = 6 kg.m². Determine:
Q2 (a)
(i)
The magnitude and direction of the torque produced on the cylinder.
(ii) The angular acceleration a of the cylinder.

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