Page 135 - Schaum's Outline of Theory and Problems of Applied Physics

120 ROTATIONAL MOTION [CHAP. 10

The potential energy of the cylinder at the top of the plane is PE = mgh. At the foot of the plane it has both
1
2
2
1
the kinetic energy of translation mv and the kinetic energy of rotation Iω , and its total KE then is equal to its
2 2
initial PE:
PE = KE
2
1
1
mgh = mv + Iω 2
2 2
Because the cylinder rolls without slipping, its linear and angular velocities are related by v = ωR, so ω = v/R.
1
2
The moment of inertia of a solid cylinder is I = mR . Hence
2
2
v

1 Iω = ( mR ) = mv 2
1
1 1
2
2
2 2 2 R 4
and the energy equation becomes
2
1
3
2
1
mgh = mv + mv = mv 2
2 4 4
Thus two-thirds of the cylinder’s KE resides in its translational motion and one-third in its rotational motion. Solving
for v yields

4 4
2
v = gh = (9.8 m/s )(1.2m) = 3.96 m/s
3 3
SOLVED PROBLEM 10.17
Find the minimum horsepower needed for the motor of the elevator of Prob. 10.13.
The required power can be calculated both from P = Fv and from P = τω:
4
P = Fv = (4000 lb)(8.33 ft/s) = 3.34 × 10 ft·lb/s
4
P = τω = (8000 lb·ft)(4.17 rad/s) = 3.34 × 10 ft·lb/s
Since 1 hp = 550 ft·lb/s,
4
3.34 × 10 ft·lb/s
P = = 60.6hp
(550 ft·lb/s)/hp
SOLVED PROBLEM 10.18
A marine diesel engine develops 90 kW at 2000 rev/min. (a) How much torque can it exert at this velocity?
(b) The engine delivers its power through a spur gear 23 cm in radius. If two teeth of the gear transmit
torque to another gear at a certain moment, find the force on each gear tooth.
(a) The angular velocity here is

2π rad/rev
ω = (2000 rev/min) = 209 rad/s
60 s/min
Hence the torque is
P 90,000 W
τ = = = 431 N·m
ω 210 rad/s
(b) Since τ = Fr, the total force is
τ 429 N·m
F = = = 1874 N
r 0.23 m
Each of the two gear teeth in contact with the driven gear exerts a force of half this amount, or 932 N.