Engineering Mechanics - PKRB: Impulse and Momentum - Discussion
Discussion Forum : PKRB: Impulse and Momentum - General Questions (Q.No. 10)
10.
A horizontal circular platform has a weight of 300 lb and a radius of gyration about the z axis passing through its center O of kz = 8 ft. The platform is free to rotate about the z axis and is initially at rest. A man, having a weight of 150 lb, begins to run along the edge in a circular path of radius 10 ft. If he has a speed of 4 ft/s and maintains this speed relative to the platform, compute the angular velocity of the platform.
= 0.313 rad/s = 0.1754 rad/s = 0.225 rad/s = 1.429 rad/sDiscussion:
1 comments Page 1 of 1.
Venkat said:
1 decade ago
v_m = v_p + v_m/p.
v_m= are x ang. Velocity + v_man.
v_m= -10 x Ang. Velocity + 4.
(H_z) _1 = (H_z) _2.
(-300 lb/g) ( (k_z) ^2) (ang.velocity) = (150/g) (v_m) (10).
(-300 lb/32.2 ft/s^2) (8^2) (ang.velocity).
= (150/32.2)(-10 x ang.velocity+4)(10).
Solve to get angular velocity = 0.175 rad/s.
v_m= are x ang. Velocity + v_man.
v_m= -10 x Ang. Velocity + 4.
(H_z) _1 = (H_z) _2.
(-300 lb/g) ( (k_z) ^2) (ang.velocity) = (150/g) (v_m) (10).
(-300 lb/32.2 ft/s^2) (8^2) (ang.velocity).
= (150/32.2)(-10 x ang.velocity+4)(10).
Solve to get angular velocity = 0.175 rad/s.
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