Engineering Mechanics - PKRB: Work and Energy - Discussion

Discussion :: PKRB: Work and Energy - General Questions (Q.No.4)


A chain that has a negligible mass is draped over a sprocket which has a mass of 2 kg and a radius of gyration of kO = 50 mm. If the 4-kg block A is released from rest in the position shown, s = 1 m, determine the angular velocity which the chain imparts th the sprocket when s = 2 m.

[A]. = 44.3 rad/s
[B]. = 39.6 rad/s
[C]. = 41.8 rad/s
[D]. = 59.1 rad/s

Answer: Option C


No answer description available for this question.

Paul said: (Feb 7, 2015)  
I am not sure if I am missing something with this question. I use the potential energy from the mass to determine the kinetic energy imparted on the sprocket. I use 4kg for the potential energy and then use 2 kg for the kinetic energy imparted on the sprocket.

From there I get a velocity and simply divide by the radius of 0.05 m to find the angular velocity but I don't get answer C.

Gege said: (Jun 9, 2017)  
T2= 1/2mv^2+1/2Iw^2,
= 1/2(4)(wr)^2+1/2(0.01)w^2,
= 0.01w^2.

T1+V1= T2+V2.
0+ 19.62N = 0.01w^2 + 0.
w = 44.29 rad/s.

Ivan said: (Dec 4, 2017)  
T2= 1/2mv^2+1/2Iw^2,
= 1/2(4)(w*0.1)^2+1/2(0.005)w^2,
= 0.0225w^2.
w=41.76 rad/s.

Mohd Farihsan Bin Zamzuri said: (May 12, 2020)  
If the angular velocity is too high at the instant s= 2m, revise the set up to reduce the angular velocity without changing the mass of block A.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.