A uniform solid sphere (I =2/5mr^2 of mass)1.5 kg and radius r = 0.548 m, is pla
ID: 1913872 • Letter: A
Question
A uniform solid sphere (I =2/5mr^2 of mass)1.5 kg and radius r = 0.548 m, is placed on the inside surface of a hemispherical bowl of radius R =3.49 m. The sphere is released from rest at an angle ? = 32.3?from the vertical and rolls without slipping (see A uniform solid sphere (I =2/5mr^2 of mass)1.5 kg and radius r = 0.548 m, is placed on the inside surface of a hemispherical bowl of radius R =3.49 m. The sphere is released from rest at an angle ? = 32.3?from the vertical and rolls without slipping (see the ?gure). The acceleration of gravity is 9.8 m/s^2. 1. How much potential energy has the sphere lost when it reaches the bottom of the bowl? Answer in units of J 2. What is the translational velocity of the sphere when it reaches the bottom of the bowl? Answer in units of m/s 3. What is the angular velocity of the sphere when it reaches the bottom of the bowl? Answer in units of rad/s
A uniform solid sphere (I =2/5mr^2 of mass)1.5 kg and radius r = 0.548 m, is placed on the inside surface of a hemispherical bowl of radius R =3.49 m. The sphere is released from rest at an angle ? = 32.3?from the vertical and rolls without slipping (see A uniform solid sphere (I =2/5mr^2 of mass)1.5 kg and radius r = 0.548 m, is placed on the inside surface of a hemispherical bowl of radius R =3.49 m. The sphere is released from rest at an angle ? = 32.3?from the vertical and rolls without slipping (see the ?gure). The acceleration of gravity is 9.8 m/s^2. 1. How much potential energy has the sphere lost when it reaches the bottom of the bowl? Answer in units of J 2. What is the translational velocity of the sphere when it reaches the bottom of the bowl? Answer in units of m/s 3. What is the angular velocity of the sphere when it reaches the bottom of the bowl? Answer in units of rad/sExplanation / Answer
1)
h=R-(R-r)sin(90-) = 3.49-(3.49-.548)sin(90-32.3) = 1.0032
potential energy lost = mgh
=1.5*9.8*1.0032 = 14.75 J
2)
conserving energy:
1/2mv^2 + 1/2 I^2 = mgh
.5*(r)^2 + .5*2/5r^2*^2 = gh
.5(r)^2 + 1/5(r)^2 = gh
r = sqrt(10/7*gh)
hence =6.84 rad/s
3)
v=r
= 6.84*.548 = 3.75 m/sec
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