Download August 29, 2007

October 26, 2009
PHY2048 Discussion-Fall ‘09
Quiz 8
Name:
UFID:
In the figure below right, a uniform solid ball of mass 0.50 g will roll smoothly along a loop-the-loop
track when released from rest along the straight section. The circular loop has radius R = 20.0 cm,
and the ball has radius r<<R.
a) What is h if the ball is on the verge of leaving the track when it
reaches the top of the loop?
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)Iω2 + mg(2R) = (7/10)mv2 + 2mgR = (7/10)mRg+2Rg = (27/10)mRg
⇒ h = (27/10)R = 2.7×20 = 54.0 cm,
where we used I = (2/5)mr2 and ω = v/r.
b) If the ball is released at height h = 5.00 R, what are the magnitude and direction of the
horizontal force component acting on the ball at point Q?
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)Iω2 + mgR = (7/10)mv2 +mgR
⇒ v = √[(10/7)g(h-R)] = √[(10/7)×9.8×0.8] = 3.35 m/s
Applying Newton’s 2nd law, the horizontal component of the force acting on the ball is
N = m(v2/R) = 5×10-4×(3.352/0.2) = 2.81×10-2 N