Question

A
2.9 kg solid sphere (radius = 0.15 m) is released from rest at the
top of a ramp and allowed to roll without slipping. The ramp is
0.85 m high and 5.2 m long.

1. When the sphere reaches the bottom of the ramp, what are
its total kinetic energy,

2. When the sphere reaches the bottom of the ramp, what is its
rotational kinetic energy?

3. When the sphere reaches the bottom of the ramp, what is its
translational kinetic energy?

Answer #1

A 1.5 kg solid sphere (radius = 0.15 m ) is released from rest
at the top of a ramp and allowed to roll without slipping. The ramp
is 0.70 m high and 5.4 m long.
When the sphere reaches the bottom of the ramp, what is its
total kinetic energy?
When the sphere reaches the bottom of the ramp, what is its
rotational kinetic energy?
When the sphere reaches the bottom of the ramp, what is its
translational kinetic...

A 2.8 kg solid cylinder (radius = 0.20 m , length = 0.50 m ) is
released from rest at the top of a ramp and allowed to roll without
slipping. The ramp is 0.80 m high and 5.0 m long.
A) When the cylinder reaches the bottom of the ramp, what is its
total kinetic energy?
B) When the cylinder reaches the bottom of the ramp, what is its
rotational kinetic energy?
C) When the cylinder reaches the bottom...

A 2.5 kg solid cylinder (radius = 0.10 m , length = 0.70 m ) is
released from rest at the top of a ramp and allowed to roll without
slipping. The ramp is 0.90 m high and 5.0 m long.
Part A When the cylinder reaches the bottom of
the ramp, what is its total kinetic energy? (Express your answer
using two significant figures.)
(answer in J)
Part B When the cylinder reaches the bottom of
the ramp, what...

A sphere of radius r0 = 23.0 cm and mass m = 1.20kg starts from
rest and rolls without slipping down a 35.0 ∘ incline that is 13.0
m long.
A. Calculate its translational speed when it reaches the
bottom.
B. Calculate its rotational speed when it reaches the
bottom.
C. What is the ratio of translational to rotational kinetic
energy at the bottom?

An 6.90-cm-diameter, 360 g solid sphere is released from rest at
the top of a 1.80-m-long, 20.0 ∘ incline. It rolls, without
slipping, to the bottom. What is the sphere's angular velocity at
the bottom of the incline? What fraction of its kinetic energy is
rotational?

An 8.50-cm-diameter, 360 g solid sphere is released from rest at
the top of a 1.90-m-long, 15.0 ∘ incline. It rolls, without
slipping, to the bottom.
Part A
What is the sphere's angular velocity at the bottom of the
incline?
Part B
What fraction of its kinetic energy is rotational?

An 8.20 cm-diameter, 390g solid sphere is released from rest at
the top of a 2.00m long, 19.0 degree incline. It rolls, without
slipping, to the bottom.
1) What is the sphere's angular velocity at the bottom of the
incline?
2) What fraction of its kinetic energy is rotational?

A solid sphere of uniform density starts from rest and rolls
without slipping a distance of d = 4.4 m down a
θ = 22°incline. The sphere has a
mass M = 4.3 kg and a radius R
= 0.28 m.
1)Of the total kinetic energy of the sphere, what fraction is
translational?
KE
tran/KEtotal =
2)What is the translational kinetic energy of the sphere when it
reaches the bottom of the incline?
KE tran =
3)What is the translational speed...

A sphere of radius r0 = 22.0 cm and mass m = 1.20kg starts from
rest and rolls without slipping down a 38.0 degree incline that is
11.0 mm long.
A. Calculate its translational speed when it reaches the
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B. Calculate its rotational speed when it reaches the
bottom.
C. What is the ratio of translational to rotational kinetic
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D. Does your answer in part A depend on mass or radius of the
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E....

A solid sphere of radius r and mass m is released from a rest on
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without slipping with its motion continuing around a loop of radius
R<<r
A) If R=0.3h, what is the speed of the sphere when it reaches
the top of the loop? Your response must be expressed in terms of
some or all of the quantities given above and physical and
numerical constants
B)...

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