Question

A hollow sphere of radius 0.24 m, with rotational inertia I = 0.036 kg · m² about a line through its center of mass, rolls without slipping up a surface inclined at 29° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 25 J.

(a) How much of this initial kinetic energy is rotational?
_________ J

(b) What is the speed of the center of mass of the sphere at the initial position?
_________m/s

Now, the sphere moves 1.0 m up the incline from its initial position.

(c) What is its total kinetic energy now?
__________J

(d) What is the speed of its center of mass now?
__________m/s

Answer 1

for a hollow sphere I=2*m*r^2/3

the energy is PE, translational KE and rotational KE
the total KE=.5*m*v^2+.5*I*ω^2
since ω=v/r
total KE=.5*m*v^2+.5*I*v^2/r^2
plug in I
total KE=.5*m*v^2+.5*2*m*v^2/3
simplify
total KE=.5*m*v^2(1+2/3)

The tranlsational has 3/5 of the energy and rotational has 2/5

a) 25*2/5=10

25*3/5=.5*m*v^2
0.036=2*m*0.24^2/3
3*0.036/(2*0.24^2)=m
m=0.93 kg
b) sqrt(2*25*3/(5*0.93))=v
v=5.68 m/s

c) Is the 1 m on the run? So the vertical gain is 1*sin(29)

total KE=25-m*g*1*sin(29)
20.58 J

d)
sqrt(2*20.58*3/(5*0.93))=v
5.15 ms