In the figure below, a solid brass ball of mass m will roll smoothly along a loo
ID: 1790984 • Letter: I
Question
In the figure below, a solid brass ball of mass m will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular motion has radius R, and the ball has radius r R. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.) (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop? (b) If the ball is released at height 6R, what is the magnitude of the horizontal force component acting on the ball at point Q?Explanation / Answer
When the ball is on the verge of leaving the track, the normal force exerted on the ball by the track is zero. The ball is still on the circular track, thus the radial component of the acceleration is v2/R. We have
m(v2/R) = mg v2 = Rg
While the ball is rolling smoothly, the mechanical energy is conserved. Applying the energy conservation equation to the initial position and the top of the circular track, we get
mgh = (1/2)mv2 + (1/2)I2 + mg(2R) = (7/10)mv2 + 2mgR = (7/10)mRg+2mRg = (27/10)mRg
h = (27/10)mRg
where we used I = (2/5)mr2 and = v/r.
The mechanical energy is conserved during the rolling motion. We apply the energy conservation equation and get the velocity at the point Q:
mgh = (1/2)mv2 + (1/2)I2 + mgR = (7/10)mv2 +mgR
v = [(10/7)g(h-R)] = [(10/7)×g×(6R-R)] = [(10/7)×g×(5R)]
Applying Newton’s 2nd law, the horizontal component of the force acting on the ball is
N = m(v2/R) = (50/7)mg