Gravitational potential energy and rotational motion

[lindzzz]

Homework Statement

A marble is placed at the top of an inverted hemispherical bowl of radius R = 0.30 m. It starts from rest and slides down the bowl without friction. Draw a free body diagram when the marble reaches an angular position θ = 16.6°. From your FBD, sketch the approximate direction of the acceleration.
1.Calculate the radial component of the acceleration (assuming that the radius of the marble is negligible). (Hint: First find the velocity at θ = 16.6°.)
2. What is the tangential component of the acceleration when θ = 16.6°? (Hint: Use Newton's second law.)
3. What is the magnitude of the normal force when the marble loses contact with the bowl?
4. Draw a free body diagram for the angular position where the marble loses contact with the bowl. Draw the direction of the acceleration next to your FBD. What is the angle θ when the marble loses contact with the bowl?

Homework Equations

PE = mgh
KE = 1/2mv²
ΔE = ΔKE + ΔPE
F = ma
ac = v²/r

The Attempt at a Solution

1. I found the radial acceleration to be 0.817 m/s² by using ΔE = ΔKE + ΔPE and found v final. I then used ac = v²/r
2. Tangential acceleration is 2.80 m/s² which I found by using ƩFx= Nx -> ma = mgsin(16.6)
3. Normal force is 0
4. I am not sure how to find this value. I tired to find the velocity when the marble loses contact and sub that into ΔE = ΔKE + ΔPE to find the final height but I wasn't sure how to find the velocity or if this is even the correct method.

 

[Simon Bridge]

well - from your answer to 3, where does this happen? Can you make an equation relating the normal force to the angle?

 

[lindzzz]

I'm not sure how I would set that up. I had thought that perhaps the angle might be 90 since that would make the x-component of the normal force equal to zero (nx = mgsinθ). But the answer was not correct. As to what happens at this point I thought that at the point the ball leaves the acceleration will be equal to gravity, but now I am unsure.

 

[Simon Bridge]

Reality check: (check my working)
The normal force is always normal to the surface of the bowl.
Gravity always points down.
Measure angle A from the top of the bowl.

for A=0, N=0, F=mg
for 0 < A < pi, F²+N²=(mg)²

so: F=mg.cos(A) and N=mg.sin(A) for the magnitudes of the tangential and normal forces. The motion is circular but not uniform - there is an angular acceleration - it's not even kinematic.

By conservation of energy, the kinetic energy will be:

T=mgR.sin(A) where R is the radius of the bowl, which puts the tangential speed at:

v²=2gR.sin(A)

centripetal acceleration is, therefore, a_c=2gsin(A), giving a centripetal force of:

F_c = 2mgsin(A)

this will be the force pointing towards the center ... which is also what the normal force does! This is twice the normal force - how come?

Anyway - from your own analysis: you'd want to say that the first time there was no contact was at A=0, the location of maximum push is at A=pi/2, so the ball loses contact with the wall at A=pi. (all angles in radiens)

So - draw a free body diagram which does not have a normal force ... what does it look like? Would this be what happens at A = pi?

 

[asteriatic]

Great job on your attempts so far! Let's go through each of your answers and provide some feedback.

1. Your approach to finding the radial acceleration is correct. However, it would be helpful to show your calculations and equations used in your response.

2. Your calculation for the tangential acceleration is correct. However, be sure to include units in your final answer.

3. The normal force is actually not 0 when the marble loses contact with the bowl. Remember that the normal force is the force that the surface exerts on the object to prevent it from falling through. In this case, when the marble loses contact with the bowl, the normal force becomes 0. To find the magnitude of the normal force, you can use Newton's second law, F=ma, and plug in the values for mass and acceleration that you have calculated.

4. To find the angle θ when the marble loses contact with the bowl, you can use the conservation of energy equation, ΔE = ΔKE + ΔPE, and set the final potential energy to 0 (since the marble is at the bottom of the bowl and has no potential energy). You can then solve for the angle θ. Alternatively, you can use the equation for conservation of energy at the top of the bowl (where the marble is initially placed) and set the final kinetic energy to 0 (since the marble has no velocity at this point). Again, you can solve for the angle θ using this equation.

Overall, your approach and understanding of the concepts involved in this problem are good. Just be sure to show all of your calculations and units in your responses, and double check your answers to ensure they make physical sense. Keep up the good work!