Home » Ask & Discuss » Physics. » Mechanics « Back to Discussion

Mechanics

Cool goIITian

Joined:	20 Nov 2008
Post:	51

27 Feb 2009 10:12:45 IST

0 People liked this

356

Prove your wisdom in Rotaional dynamics

None

A force 100 Newton is exerted on a disc which has a mass of 1Kg and radius 1m. The force is exerted on the topmost point on the dis. The coefficient of friction is 0.5. What is the total work done by friction when the disc moves a distance 1m.

Share this article on:

Comments (2)

Soumen Goswami

Forum Expert

Joined: 10 Apr 2007

Posts: 1777

7 Mar 2009 19:03:39 IST

Like 1 people liked this

The bottommost point slips backwards. So friction acts forward with its max value 5 (Assuming g = 10). The acceleartion is . The net torque is 95, so the angular accelerations . The velocity of the centre is then and angular velocity of the disk at any time is . Here is never possible. So, friction will keep on acting and the point of contact will keep on slipping. The work done is

ABINASH

Hot goIITian

Joined: 20 Apr 2007

Posts: 131

9 Mar 2009 23:54:34 IST

Like 0 people liked this

first, we approach hoping this as a rolling case.... so, first we infer that friction acts in the direction of F. So, using equations....

F + f = mA_cm and F - f = alpha/2 = A_cm/2....

we get A_cm = 4F/3 or f = F/3....

Bt, f_max = 5N. So, we conclude that sliding is there. So, f = 5N here and acts in the direction of F. So, W_f = 5J...

Hope this helps...

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team