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Rotational motion of belt drive

  • Thread starter kidsmoker
  • Start date
1. Homework Statement

I'm just working through a few past mechanics papers and this question came up:

A belt drives the circumference of a cylindrical wheel of radius R and mass M with
no slipping. The tension in the belt is T. Friction in the wheel bearing causes an
effective torque N to act on the wheel.
(i) If the velocity of the belt is constant, what is T in terms of N.
(ii) The bearing is lubricated removing all significant friction and the velocity of
the belt then increases with a uniform acceleration a. Write down T in terms of
M, R, and a.

I'd be greatful if someone could just check through my solutions so that i'm sure doing things correctly.

2. Homework Equations

I=0.5mR^2 for disc
v=wr
L=I(dw/dt)

3. The Attempt at a Solution

(i) T=N .

(ii) v=wR implies dw/dt = a/R .

Therefore TR = 0.5MR^2 * (a/R) ,

and so T = 0.5Ma .

Thanks.
 

djeitnstine

Gold Member
614
0
One thing I immediately noticed. In [tex]T \neq N[/tex] Why? because this is saying that Force is equal to torque. Surely you know the relationship between torque and force.
 
One thing I immediately noticed. In [tex]T \neq N[/tex] Why? because this is saying that Force is equal to torque. Surely you know the relationship between torque and force.


Ah yeah woops! Should be T=N/R ?

Thanks.
 

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