Rotational motion of belt drive

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SUMMARY

The discussion focuses on the mechanics of a belt drive system involving a cylindrical wheel with radius R and mass M. The tension in the belt (T) is analyzed in two scenarios: first, when the belt's velocity is constant, leading to the conclusion that T equals the torque (N) divided by the radius (R), or T = N/R. In the second scenario, where the bearing is lubricated and the belt accelerates uniformly, the tension is derived as T = 0.5Ma, where a is the acceleration. The participants clarify the relationship between tension and torque, emphasizing the correct application of mechanics principles.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with the equations of motion for rotating bodies
  • Knowledge of the relationship between linear and angular velocity
  • Basic principles of friction in mechanical systems
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  • Study the relationship between torque and force in rotational systems
  • Explore the effects of lubrication on mechanical efficiency
  • Learn about the dynamics of belt drives and their applications
  • Investigate the equations of motion for non-uniform acceleration in rotating systems
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Students and professionals in mechanical engineering, particularly those focusing on rotational mechanics and belt drive systems.

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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.

Homework Equations



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

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.
 
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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.
 
djeitnstine said:
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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