An 8.70-cm-diameter, 320g solid sphere is released from rest at the top of a 2.0

Question

An 8.70-cm-diameter, 320g solid sphere is released from rest at the top of a 2.00-m-long, 20 degree incline. It rolls, without slipping, to the bottom. A) What is the sphere's angular velocity at the bottom of the incline? B) What fraction of its kinetic energy is rotational? An 8.70-cm-diameter, 320g solid sphere is released from rest at the top of a 2.00-m-long, 20 degree incline. It rolls, without slipping, to the bottom. A) What is the sphere's angular velocity at the bottom of the incline? B) What fraction of its kinetic energy is rotational? A) What is the sphere's angular velocity at the bottom of the incline? B) What fraction of its kinetic energy is rotational?

Explanation / Answer

h=lsin thetha

=2sin20

=1.82m

k= .5I?^2
I(solid sphere)= (2/5)MR^2

Eo=Ef
Po+Ko=Pf+Kf
Po=Krot+Ktrans
mgh=.5I?^2+.5mv^2
mgh=.5(2/5)MR^2 (v/R)^2+.5mv^2
gh= (2/10)v^2 + .5v^2

v=sqrt[ (10/7)gh )

=sqrt(1.42*9.8*1.82)

=sqrt(25.3)

v=5.02m/sec
?=(v/r)=

=5.02/(0.087/2)

= 5.02/0.0435

w=115.4rad/sec

b)Now let us calculate the kinetic energy.
KEr = (1/2)I*w^2 = (1/2)[2*M*R^2/5]*w^2
KEr = (1/5)[M*R^2]*w^2
KEr = 0.2(0.320*(0.0435)^2)(115.4)^2

KEr=1.61J

Total kinetic energy = PE = M*g*H

=0.320*9.8*1.82

=5.7J

Fraction of the kinetic energy that is rotational = KEr/PE = 1.61/5.7
Fraction = 0.282

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