A solid ball rolls up a ramp (inclined plane) of h=0.50 m high withou slipping. It has an initial velocity of its center of mass of -3.0 m/s when the ball is at the bottom of the ramp. (a) What is its velocity at the top of the ramp? (b) If the ramp is 1.0 m high, does it make it to the top?
A solid ball rolls up a ramp (inclined plane) of h=0.50 m high withou slipping. It has an initial velocity of its center of mass of -3.0 m/s when the ball is at the bottom of the ramp. (a) What is its velocity at the top of the ramp? (b) If the ramp is 1.0 m high, does it make it to the top?
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter10: Fixed-axis Rotation
Section: Chapter Questions
Problem 69P: A solid sphere of radius 10 cm is allowed to rotate freely about an axis. The sphere is given a...
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![A solid ball rolls up a ramp (inclined plane) of h=0.50 m high without
slipping. It has an initial velocity of its center of mass of √-3.0 m/s
when the ball is at the bottom of the ramp.
(a) What is its velocity at the top of the ramp?
(b) If the ramp is 1.0 m high, does it make it to the top?
Vc.
0=25°
М=0.22
f
masina
VA
Vmg Cso
da
h
Sind
AB=d
Ac = D
H
Homework #2
fdesetin
a) AE (A+B) = WN₁ = -fd=setd
(K=+*+83) - (KA+Kx)=="tond
b) Consider Vc=0
DE (AC) =WNL
скат -
Pgc - (KA+K+) A = ut D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2dbe910a-0548-4393-97e4-b2099f203113%2F94badf0b-afa1-4447-8c7d-8c14be0f5310%2Fr2a71r8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A solid ball rolls up a ramp (inclined plane) of h=0.50 m high without
slipping. It has an initial velocity of its center of mass of √-3.0 m/s
when the ball is at the bottom of the ramp.
(a) What is its velocity at the top of the ramp?
(b) If the ramp is 1.0 m high, does it make it to the top?
Vc.
0=25°
М=0.22
f
masina
VA
Vmg Cso
da
h
Sind
AB=d
Ac = D
H
Homework #2
fdesetin
a) AE (A+B) = WN₁ = -fd=setd
(K=+*+83) - (KA+Kx)=="tond
b) Consider Vc=0
DE (AC) =WNL
скат -
Pgc - (KA+K+) A = ut D
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