Solving Frames of Reference Homework: Static/Kinetic Friction

[JWSiow]

Homework Statement

A child sits 2m from the centre of a merry-go-round (a rotating wooden platform) which is rotating at 3.5revs/min in a clockwise direction (when viewed from above). She places a wooden block of mass m=0.25kg beside her on the platform.

a)Draw a diagram showing all the forces acting on the block, and describe each force.

b)If the block remains at rest relative to the platform, calculate its acceleration. Explain your reasoning.

c)If the coefficients of static and kinetic friction between the block and the platform are 0.2 and 0.18 respectively, do you expect the block to remain at rest or to slide relative to the platform?

Homework Equations

F(friction) = $$\mu$$_kF_N
F(friction) = $$\mu$$_sF_N

The Attempt at a Solution

a)I'm not sure if I had the write forces in my diagram, I had the weigh mg, normal force, centripetal force and friction force.

b)I'm not sure about this one either, but I think, since the frame of reference is the platform, and the block is not moving, acceleration = 0. ?

c)I didn't really know what to do for this, so I worked out the static and kinetic friction forces, and then compared it to the centripetal force. Since the centripetal force was smaller than the static friction force, the block would remain at rest. I don't know if this is correct, or if it's even a good explanation!

 

[Bartek]

a)I'm not sure if I had the write forces in my diagram, I had the weigh mg, normal force, centripetal force and friction force.

b)I'm not sure about this one either, but I think, since the frame of reference is the platform, and the block is not moving, acceleration = 0. ?

c)I didn't really know what to do for this, so I worked out the static and kinetic friction forces, and then compared it to the centripetal force. Since the centripetal force was smaller than the static friction force, the block would remain at rest. I don't know if this is correct, or if it's even a good explanation!

(a) friction force is the centripetal force.,so you have three forces ("centripetal force" is just the name of real force directed to the center. In this case - friction force)

(b) No. Platform is not inertial frame of reference. So in this frame of reference there is centrifugal force. You have two possibilities: use ground frame of refference and calculate centripetal acceleration (recommended ) OR use platform as a frame of refference and calculate centrifugal acceleration.

(c) OK, but block would remain rest since centripetal force was smaller OR EQUAL to static friction force

regards

 

[JWSiow]

Ok, thanks. :)

So, for b), when calculating the centripetal acceleration, I'd use the velocity of the platform?

 

[Bartek]

So, for b), when calculating the centripetal acceleration, I'd use the velocity of the platform?

Yes. And distance between mass and pivot. Find centripetal force first.

regards

 

[rwisz]

I would like to provide some clarification and suggestions for the above solution attempt.

a) The diagram should include the following forces acting on the block: weight (mg), normal force (FN), centripetal force (Fc), and friction force (Ff). It is important to label each force and indicate its direction on the diagram. The weight of the block is pulling it downwards, the normal force is perpendicular to the surface of the platform, the centripetal force is directed towards the center of the merry-go-round, and the friction force is parallel to the surface and opposite to the direction of motion.

b) Since the block is at rest relative to the platform, its acceleration will be zero. This is because the net force acting on the block is also zero, according to Newton's first law of motion. In this case, the normal force and the friction force are balancing the weight and the centripetal force, resulting in zero net force and zero acceleration.

c) To determine if the block will remain at rest or slide, we need to compare the maximum static friction force (Fmax = \musFN) with the applied force (Fc). If the applied force is less than the maximum static friction force, the block will remain at rest. However, if the applied force is greater than the maximum static friction force, the block will start to slide and the kinetic friction force (Fk = \mukFN) will come into play. In this case, since the applied force (Fc) is smaller than the maximum static friction force, the block will remain at rest.

Overall, it is important to clearly define the frame of reference and carefully consider all the forces acting on the block in order to accurately solve this problem. It is also helpful to clearly label all the forces and explain the reasoning behind each step in the solution.