Solving Drive Train Problem with Friction Coefficient of 0.3

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This discussion focuses on solving a drive train problem involving a wheel with a mass of 10 kg and a radius of 10 cm, driven by a motor with a torque of 5 Nm. The key calculations include determining the time to accelerate the wheel to 100 rpm and analyzing the tensions in the belt given a friction coefficient of 0.3. The participants confirm that the torque from the motor equals the moment of inertia times angular acceleration, and that tension in a massless belt is uniform unless friction is considered, which complicates the analysis.

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  • Newtonian Mechanics principles
  • Understanding of torque and angular acceleration
  • Basic concepts of friction and tension in belts
  • Moment of inertia calculations
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  • Calculate angular acceleration using torque and moment of inertia
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Mechanical engineers, physics students, and anyone involved in the design or analysis of drive train systems and mechanical systems involving friction and tension.

ee7klt
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Hi all,
I was wondering if someone could help me out on this (taken from Newtonian Mechanics, A.P. French, pg 705, problem14-23)

A wheel of uniform thickness , of mass 10kg and radius 10cm is driven by a motor through a belt. The drive wheel on the motor is 2cm in radius. The motor is capable of delivering a torque of 5 Nm.

A) Assuming that the belt does not slip on the wheel, how long does it take to accelerate the large wheel from rest up to 100 rpm?

B) If the coefficient of friction between belt and wheel is 0.3, what are the tensions in the belt on the two sides of the wheel?
(Assume that the belt touches the wheel over half its circumference.)

For A, I had that if the belt does not slip, all of the torque is transferred to the driven wheel and I applied torque = moment X angular acceleration. And alpha = omega X time assuming alpha is constant.

For B, I thought of deriving friction from the normal force exerted by circumference on belt, but this looks like a nightmare to compute since the force the belt exerts on the circumference varies over the circumference.

In any case, does not the tension In a massless rope have to the the same on both sides always? If not we get infinite acceleration in any given differential element!

thanks.
 
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ee7klt said:
In any case, does not the tension In a massless rope have to the the same on both sides always? If not we get infinite acceleration in any given differential element!

thanks.

Welcome to PF ee7klt!

For a rotating massless pulley, [tex]\alpha \neq 0[/tex] but moment of inertia is zero as mass is zero. Hence, the tensions have to be same on either side of the pulley (you can see why this is so by making a freebody diagram of a pulley whose center of mass is attached to a rope the tension in which is T and which is wrapped by an inextensible rope--now find the tensions in either part of the rope using the idea just mentioned).

If however, friction exists, the analysis changes a bit (you have to consider an element of the rope), you have to use a relation involving the (different) tensions, T1 and T2 on either side of the pulley, the coefficient of static friction of the pulley-rope interface and the angle of wrap. Think about this.
 
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Hi there,

For part A, your approach is correct. The torque exerted by the motor is equal to the moment of inertia of the wheel times its angular acceleration. Since the wheel starts from rest, its initial angular velocity is 0 and we can solve for the angular acceleration. Then, using the formula for angular velocity, we can calculate the time it takes to reach 100 rpm.

For part B, you are correct that the tension in a massless rope or belt should be the same on both sides. This is because the rope or belt is considered to be infinitely thin and therefore does not have any mass to affect the tension. However, the tension in the belt will vary along its length due to the varying friction force between the belt and the wheel. To solve for the tensions on either side of the wheel, you can consider the forces acting on the belt (tension, normal force, and friction force) and use the coefficient of friction to calculate the friction force. Then, using the fact that the net force on the belt must be equal to its mass times its acceleration, you can solve for the tensions on either side of the wheel.

I hope this helps. Good luck with your problem!
 

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