Use conservation of mechanical energy (potential, linear and rotational kinetic

Question

Use conservation of mechanical energy (potential, linear and rotational kinetic energy) and set the centripetal force equal to the weight to find the minimun height.

If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it at point Q?

A solid marble of mass m = 10 kg and radius r = 2 cm will roll without slipping along the loop-the-loop track shown in the figure if it is released from rest somewhere on the straight section of track. The radius of the loop-the-loop is R = 1.35 m. From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? Use conservation of mechanical energy (potential, linear and rotational kinetic energy) and set the centripetal force equal to the weight to find the minimun height. If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it at point Q?

Explanation / Answer

A.) If we look at the top of the loop, what it is telling us is that the normal between the marble and the ramp is zero for just an instant. If we sum the forces at this point, the only force acting on the marble is gravity. In this situation, gravity is pointing into the center of the arc. That means we can can sum the forces and call our acceleration centripital acceleration. Through a proof I'm not going to go through, centripital accerlation can be related to tangental velocity and the radius it is acting at.

Ac=V^2/R
Sum of the forces..
m*V^2/R=m*g

Solve for V^2
=g*R

Now we can relate velocity to high dropped by calling our system the marble, ramp, and the earth. If this is our system, there is no external work done, so the change in energy in the system is equal to zero.

0= change in kinetic energy plus change in gravitational potential.

0=(KEf-KEi)+(mgHf-mgHi)
Lets say that at the top of the loop, h=0. KEi=zero, so now we can write

KEf=mgHi
(1/2)mV^2=mgHi
sub in V^2
m*g*R/2=mgHi

Hi=R/2

Therefore the height total is equal to R/2+2R, or 3R/2. Recall that I was doing this in terms from the top of the loop, so we need to add back in the diameter of the loop.

B.) So the total change in height is 5R moving to point Q. Using change in energy, we can deduce the velocity at the point.

Vf^2=10g*R

At the point Q, the normal from the ramp on the marble is directed into the center of the arc. Again, Ac=V^2/R
Sum the forces in the x direction,

m*Ac=Fnormal
m*V^2/R=Fnormal
m*10*g=Fnormal

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