Torque of a rotating wheel on a static wheel

In summary, a uniform cylindrical wheel of mass ##m_{1}## and radius ##R_{1}## rotating with angular velocity ##\omega_{1}## exerts a torque of ##\frac{2}{3}R_{1}\mu F## on a static wheel of radius ##R_{2}## and mass ##m_{2}## when they are pushed against each other with a constant force ##F## distributed evenly across the wheel's face. This torque is in the same direction as the wheel's angular velocity. The friction between the wheels is taken into account through the coefficient of friction ##\mu## and follows Newton's Third Law, where the reaction torque is equal in magnitude but opposite in direction.
  • #1
Jenny Physics
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4

Homework Statement


A uniform cylindrical wheel of mass ##m_{1}## and radius ##R_{1}## rotates with angular velocity ##\omega_{1}##. It lies a certain distance (along the same axis) from a static wheel of radius ##R_{2}## and mass ##m_{2}##. The wheels are then pushed against each other with a constant force ##F## uniformly distributed across the wheel's face. There is friction ##\mu## between the wheels.
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When the wheels first touch what is the magnitude of the torque that wheel 1 exerts on wheel 2 via friction?

Homework Equations


Split wheel 1 into infinitesimal concentric rings of radius ##r## and width ##dr## and find the torque exerted by one of these then integrate to find the total torque.

The Attempt at a Solution



I probably should use ##d\tau=dI \alpha## where ##dI## is the moment of inertia due to the cylindrical ring. But since ##\omega_{1}=const## it would seem that ##\alpha=\dot{\omega}_{1}=0##?
 

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  • #2
Jenny Physics said:

Homework Statement


A uniform cylindrical wheel of mass ##m_{1}## and radius ##R_{1}## rotates with angular velocity ##\omega_{1}##. It lies a certain distance (along the same axis) from a static wheel of radius ##R_{2}## and mass ##m_{2}##. The wheels are then pushed against each other with a constant force ##F## uniformly distributed across the wheel's face. There is friction ##\mu## between the wheels.
View attachment 232622
When the wheels first touch what is the magnitude of the torque that wheel 1 exerts on wheel 2 via friction?

Homework Equations


Split wheel 1 into infinitesimal concentric rings of radius ##r## and width ##dr## and find the torque exerted by one of these then integrate to find the total torque.

The Attempt at a Solution



I probably should use ##d\tau=dI \alpha## where ##dI## is the moment of inertia due to the cylindrical ring. But since ##\omega_{1}=const## it would seem that ##\alpha=\dot{\omega}_{1}=0##?

You are interested in torque exerted when they touch. The normal force pushing them together will cause a tangential force through friction that will exert the torque. Thing about using ##\tau = r \times F##.
 
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  • #3
Dick said:
You are interested in torque exerted when they touch. The normal force pushing them together will cause a tangential force through friction that will exert the torque. Thing about using ##\tau = r \times F##.
Using ##d\tau=r\frac{F}{\pi R_{1}^{2}}\pi rdr## and then ##\tau=\int_{0}^{R_{1}}d\tau##?
 
  • #4
Jenny Physics said:
Using ##d\tau=r\frac{F}{\pi R_{1}^{2}}\pi rdr## and then ##\tau=\int_{0}^{R_{1}}d\tau##?
Close. The surface of a ring of thickness ##dr## and radius ##r## is ##dA=2\pi r dr##.
 
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  • #5
kuruman said:
Close. The surface of a ring of thickness ##dr## and radius ##r## is ##dA=2\pi r dr##.

And ##F## is the normal force. The force acting against the rotation is the frictional force. There is a ##\mu## involved!
 
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  • #6
kuruman said:
Close. The surface of a ring of thickness ##dr## and radius ##r## is ##dA=2\pi r dr##.
Then ##d\tau=r\frac{F}{\pi R_{1}^{2}}2\pi rdr=2\frac{F}{R^{2}_{1}}r^{2}dr## and then ##\tau=\int_{0}^{R_{1}}d\tau=\frac{2}{3}R_{1}F##? The torque will be tangential to the surface and counterclockwise?
 
  • #7
Dick said:
And ##F## is the normal force. The force acting against the rotation is the frictional force. There is a ##\mu## involved!
The friction force torque will then be ##\tau=-\frac{2\mu}{3}R_{1}F_{1}## so that the total torque will be ##\tau=\frac{2}{3}R_{1}F_{1}(1-\mu)## in the same direction as ##\omega_{1}##?
 
  • #8
Jenny Physics said:
The friction force torque will then be ##\tau=-\frac{2\mu}{3}R_{1}F_{1}## so that the total torque will be ##\tau=\frac{2}{3}R_{1}F_{1}(1-\mu)## in the same direction as ##\omega_{1}##?
Yes. Since the angular speed of the wheel is increasing, the torque on it and its angular velocity must be in the same direction.

Where does ##(1-\mu)## come from? What happens when ##\mu = 1##? Check your integral.
 
  • #9
kuruman said:
Yes. Since the angular speed of the wheel is increasing, the torque on it and its angular velocity must be in the same direction.

Where does ##(1-\mu)## come from? What happens when ##\mu = 1##? Check your integral.
Not following. The ##(1-\mu)## comes from the torque of ##F## and the opposite torque of the friction force which is ##\mu F##. Am I missing ##2\pi## in the integral?
 
  • #10
Newton's 3rd law says that to every action there is an equal and opposite reaction. In other words, you change the sign but not the magnitude. This applies to forces and, since torques are generated by forces, it applies to torques as well.
 
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  • #11
Jenny Physics said:
Not following. The ##(1-\mu)## comes from the torque of ##F## and the opposite torque of the friction force which is ##\mu F##. Am I missing ##2\pi## in the integral?

##F## doesn't produce any net torque, contributions from opposite sides of the wheel cancel. Only the frictional force produces a net torque.
 
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  • #12
Dick said:
##F## doesn't produce any net torque, contributions from opposite sides of the wheel cancel. Only the frictional force produces a net torque.
The torque will then be ##\frac{4\pi}{3}R_{1}\mu F##?
 
  • #13
Jenny Physics said:
The torque will then be ##\frac{4\pi}{3}R_{1}\mu F##?

Where did the '4' come from?
 
  • #14
Dick said:
Where did the '4' come from?
Kuruman mentioned my integral was missing something I thought I was missing a ##2\pi##. Perhaps he meant the factor ##1-\mu## should just be ##\mu##?
 
  • #15
Jenny Physics said:
Kuruman mentioned my integral was missing something I thought I was missing a ##2\pi##. Perhaps he meant the factor ##1-\mu## should just be ##\mu##?

You are throwing factors in because you are guessing. Go over it from the beginning and show what YOU think the result should be.
 
  • #16
Dick said:
You are throwing factors in because you are guessing. Go over it from the beginning and show what YOU think the result should be.
I think it should be ##\frac{2}{3}R_{1}\mu F##
 
  • #17
Jenny Physics said:
I think it should be ##\frac{2}{3}R_{1}\mu F##

And I think that's correct.
 
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  • #18
Jenny Physics said:
Kuruman mentioned my integral was missing something I thought I was missing a ##2\pi##. Perhaps he meant the factor ##1-\mu## should just be ##\mu##?
Yes, not knowing where the ##(1-\mu)## came from, I questioned whether it came as a result of faulty integration.
 
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FAQ: Torque of a rotating wheel on a static wheel

What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is typically measured in units of Newton-meters (N·m) or pound-feet (lb·ft).

How is torque calculated?

Torque is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied. The formula for torque is T = F x r, where T is torque, F is force, and r is the distance from the axis of rotation to the point where the force is applied.

What is the equation for torque of a rotating wheel on a static wheel?

The equation for torque of a rotating wheel on a static wheel is T = I x α, where T is torque, I is the moment of inertia, and α is the angular acceleration.

How does the torque of a rotating wheel on a static wheel affect its speed?

The torque of a rotating wheel on a static wheel is directly proportional to its angular acceleration. This means that a larger torque will result in a faster increase in speed, while a smaller torque will result in a slower increase in speed.

What factors can affect the torque of a rotating wheel on a static wheel?

The torque of a rotating wheel on a static wheel can be affected by the force applied, the distance from the axis of rotation, and the moment of inertia of the wheel. Additionally, external factors such as friction and air resistance can also affect the torque and speed of the rotating wheel.

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