Fall 2015, Physics 2701. Lab 9. Challenge Lab 9. Challenge Lab Pre-Laboratory Ho

Question

Fall 2015, Physics 2701. Lab 9. Challenge Lab 9. Challenge Lab Pre-Laboratory Homework Assignment Read Knight's textbook, Chapter 10.3(Kinetic Energy) and 10.4 (Potential Energy) Solve the following problem using the Prepare-Solve-Assess technique: A block of mass m is released from rest at a height h. The block slides down the ramp without any friction or air resistance and then passes along a circular loop of radius R. Given m, g, and R find the minimum height h so that the block never loses contact with the track/loop.

Explanation / Answer

For minimum value of h, the car is just able to remain in contact with the track at the top of the loop.
Therefore, force of gravity is the only force that provides centripetal acceleration.

At the top of the loop: -
Force of gravity = mg vertically downward
Centripetal force = mv^2/R vertically downward (where v = car's speed)

Therefore mg = mv^2/R
mg = 2*(1/2 mv^2)/R
mg = 2*(K.E)/R -------------1

Now,
Let
K1 = kinetic energy of the car at the height h,
U1 = gravitational potential energy of the car at the height h,
K2 = kinetic energy of the car at the top of the loop,
U2 = gravitational potential energy of the car at the top of the loop

As The track is frictionless. Therefore, mechanical energy is conserved.
Therefore,
K1 + U1 = K2 + U2
0 + mgh = K2 + 2mgR
K2 = mg(h - 2R)

Substituing Value into eq1
mg = 2*mg(h - 2R)/R
1 = 2(h-2R)/R
R = 2(h - 2R)
2h = 5R
h = (5/2)R

Minimum Height, h = (5/2)R

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