Physics HW Help: Find Force for Wheat on Conveyor Belt

[cwgeary]

Any help would be appreciated. I have the answers to these problems but can't seem to fully work through them. Well here they are:

#1) A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction iwth the turntable surface of static friction=0.80 and kinetic friction=0.50. The turntable very slowly speeds up to 60 rpm. Does the coin slide off?

The answer is no. But I am not sure how to get there. I converted 60 rpm to radians per second and figured the maximum static friction before the coin will move is 0.0392 (0.80*0.049). I tried to figure what the force on the penny would be by using Newton's 2nd law: sum of the forces in the r direction=mass*angular velocity*radius (0.005*(2*PI)^2)(0.15). Is any of this right?

#2) Wheat is poured onto a convyor belt at a steady rate of 30 kg/s. What force must be exerted to keep the belt moving at a constant speed of 2.0 m/s? Any help on this one would be appreciated. I am not sure how to even start attacking this one.

 

[hotvette]

https://www.physicsforums.com/showthread.php?t=94379

 

[quicknote]

Sure, I'd be happy to help with these problems. Let's tackle them one at a time.

For problem #1, you're definitely on the right track. To determine whether the coin will slide off the turntable, we need to compare the maximum static friction force with the force required to keep the coin in circular motion. Let's break it down step by step:

1. Convert the rotational speed of 60 rpm to radians per second. We know that 1 revolution = 2π radians, so 60 rpm = 2π*60/60 = 2π radians per second.

2. Use this value to calculate the force required to keep the coin in circular motion. According to Newton's 2nd law, the force in the radial direction (towards the center of the turntable) would be equal to the mass of the coin times the square of the angular velocity times the radius. So, F = (0.005 kg)*(2π rad/s)^2 * (0.15 m) = 0.0471 N.

3. Compare this force with the maximum static friction force, which is equal to the coefficient of static friction (0.80) times the normal force between the coin and the turntable. The normal force is just the weight of the coin, which is equal to mg = (0.005 kg)*(9.8 m/s^2) = 0.049 N. So, the maximum static friction force is 0.80*0.049 N = 0.0392 N.

4. Since the force required to keep the coin in circular motion (0.0471 N) is greater than the maximum static friction force (0.0392 N), the coin will not slide off the turntable. In other words, the turntable is able to provide enough friction to keep the coin in place.

For problem #2, we can use a similar approach to determine the force required to keep the conveyor belt moving at a constant speed. Here's how we can do it:

1. First, let's calculate the weight of the wheat that is being poured onto the conveyor belt per second. We know that the rate of pouring is 30 kg/s, so the weight of the wheat per second would be 30 kg/s * 9.8 m/s^2 = 294 N/s.

2. Now, let's think about the forces acting on the wheat on the