Rolling of a disc

  • #1

Homework Statement



A cart with mass M has four wheels (idealized as uniform discs), each of radius r and mass m, arranged symmetrically with respect to the cart. Find the acceleration of the cart when a horizontal force F is applied on it. There is no slipping between the wheels and the horizontal road.

2. Homework Equations

a = rα
Γ = Iα
(M+4m)a = F-4f.

The Attempt at a Solution



I first calculated the linear acceleration w.r.t. ICM using a = rα. The net torque on any disc will be due to frictional force and will be equal to f*r (f=frictional force). I equated this to Iα and then finally used (M+4m)a = F-4f to calculate the value of a.
In this problem, the wheels are rolling on the ground without slipping. In that case, shouldn't the friction on the lower most part be equal to zero. Why are we taking friction into account? Also, assuming we take friction into account, was what I did above correct?

Edit: Is there friction there in order to prevent slipping?
 
Last edited:

Answers and Replies

  • #2
173
29
Forget friction , there's no mention of any .... the applied force is going into accelerating the cart (I take cart to mean the body and 4 wheels )..so the whole mass of the cart including wheels is given an acceleration , but also some force is spent in turning the 4 wheels .
 
  • #3
ehild
Homework Helper
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Homework Statement



A cart with mass M has four wheels (idealized as uniform discs), each of radius r and mass m, arranged symmetrically with respect to the cart. Find the acceleration of the cart when a horizontal force F is applied on it. There is no slipping between the wheels and the horizontal road.

2. Homework Equations

a = rα
Γ = Iα
(M+4m)a = F-4f.

The Attempt at a Solution



I first calculated the linear acceleration w.r.t. ICM using a = rα. The net torque on any disc will be due to frictional force and will be equal to f*r (f=frictional force). I equated this to Iα and then finally used (M+4m)a = F-4f to calculate the value of a.
In this problem, the wheels are rolling on the ground without slipping. In that case, shouldn't the friction on the lower most part be equal to zero. Why are we taking friction into account? Also, assuming we take friction into account, was what I did above correct?

Edit: Is there friction there in order to prevent slipping?

Yes, there must be friction to prevent slipping. This friction is static.
You are on the right track, go ahead.
 
  • #4
ehild
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Forget friction , there's no mention of any ...
The problem says that the wheels roll on the ground without slipping. Is it possible without friction between the ground and the wheels?
 
  • #5
Yes, there must be friction to prevent slipping. This friction is static.
You are on the right track, go ahead.
Thanks :D
 
  • #6
173
29
The problem says that the wheels roll on the ground without slipping. Is it possible without friction between the ground and the wheels?

When I said "forget friction" ... I meant ignore it , it doesn't affect the calculation , no force is used up in overcoming friction .... no energy is lost in friction
 
  • #7
ehild
Homework Helper
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When I said "forget friction" ... I meant ignore it , it doesn't affect the calculation , no force is used up in overcoming friction .... no energy is lost in friction
The force of friction adds to the applied force F, so it influences the acceleration of the cart, and its torque accelerates the rotation of the wheels. As it is static friction, its work is zero, you are right in that statement.
 
  • #8
173
29
The force of friction adds to the applied force F, so it influences the acceleration of the cart, and its torque accelerates the rotation of the wheels. As it is static friction, its work is zero, you are right in that statement.

Yes .... I tend to look at where is the force going ...in accelerating the whole mass cart and wheels in a strait line , and also in increasing the angular speed of the wheels ... Just those two places
 
  • #9
haruspex
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When I said "forget friction" ... I meant ignore it , it doesn't affect the calculation , no force is used up in overcoming friction .... no energy is lost in friction
Ok, perhaps, but it was very misleading. Please try to be more careful when offering guidance.
Force "being used up" is not a well defined concept. Whether it affects the calculation depends on the approach used. It didn't affect it in your calculation, but in other methods it could.
 

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