Question

A hollow sphere of radius 0.24 m, with rotational inertia
*I* = 0.036 kg · m^{2} 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

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