A query about torque

  • Thread starter recon
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  • #1
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The following is a problem from the examples section of my school textbook, so it is accompanied by a full solution. However, I do not fully understand it and need your guidance in clearing my doubts.

http://img151.imageshack.us/img151/7024/picture30rr.jpg [Broken]

"A cyclist is moving at speed v around a bend of radius r. The forces acting on the cyclist are N, the normal reaction due to the ground on the bicycle, mg, the force due to gravity, and f, the frictional side-thrust on the tyres of the bicycle.

For vertical equilibrium, N = mg --(1) [I agree with this.]

For circular motion, f = m(v^2)/r -- (2) [Yes, I understand this.]

Equations (1) and (2) taken together give f/N = (v^2)/rg. "

I do not understand the next section of the solution:

"The clockwise torque about the centre of gravity, G = fh cos(theta). The anticlockwise torque about the centre of gravity, G = Nh sin (theta).

For stability fh cos (theta) = Nh sin (theta), or f/N = tan (theta).

Substituting this into Equation (3), we have tan (theta) = (v^2)/rg."

I don't understand why it is possible to equate the net torque about the centre of gravity to be zero. As the cyclist is not in equilibrium, shouldn't both the net torque and the net force acting on the cyclist be more than zero? Also if we took the net torque about the point of contact of the bicycle with the ground, then it clearly isn't zero because the torque produced by mg isn't balanced out by any other force. :grumpy:
 
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Answers and Replies

  • #2
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If there is a net torque, it's angular speed should change. But it appears that the angular speed is not changing (question says it's v/r, and doesn't mention that it's changing), it just turning with a constant speed. No net force doesn't mean no motion, it means no accelearion. In short:

Newton's first law.
 
  • #3
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Can someone explain why there is a net torque about the point of contact of the bicycle with the ground?
 
  • #4
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recon said:
Can someone explain why there is a net torque about the point of contact of the bicycle with the ground?

Look at the forces in your diagram. Taking the axis of rotation to be the point where the bicycle tire meets the ground there is only one force that has a non-zero moment arm: the weight. Thus there is a net torque about that axis.

-Dan
 
  • #5
Doc Al
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recon said:
I don't understand why it is possible to equate the net torque about the centre of gravity to be zero. As the cyclist is not in equilibrium, shouldn't both the net torque and the net force acting on the cyclist be more than zero?
While the translational acceleration of the cyclist is certainly nonzero (since it's obviously centripetally accelerating), the rotational acceleration is zero. (At least to a good approximation--ignoring the angular momentum of the spinning wheels.)


Also if we took the net torque about the point of contact of the bicycle with the ground, then it clearly isn't zero because the torque produced by mg isn't balanced out by any other force.
You can treat the motion of the cycle as a combination of the motion of its center of mass plus its rotation about the center of mass. As long as you stick to that, you don't have to worry about the fact that the cycle is not an inertial frame.

But if you choose to find torques about a point other than the center of mass (such as the point of contact with the ground) you have to include the effect of the "fictitious" inertial forces (aka, centrifugal force). So, in addition to the torque produced by mg, there is a balancing torque due to the centrifugal force, which equals m(v^2)/r and acts radially outward through the center of mass. (You should convince yourself that the lean angle is the same no matter how you calculate it.)
 
  • #6
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My high school physics class does not cover non-inertial frames, but anyway I think I understand now, and I'll try and read up on it later. Thank you very much!!! :)
 

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