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A sphere of radius r =34.5 cm and mass m = 1.80 kg starts from rest...

Question

A sphere of radius r =34.5 cm and mass m = 1.80 kg starts from rest...

A sphere of radius r =34.5 cm and mass m = 1.80 kg starts from rest and rolls without slipping down a 30.0? incline that is 10.0m long.

Part A

Calculate its translational speed when it reaches the bottom.

v=

Part B

Calculate its rotational speed when it reaches the bottom.

Express your answer using three significant figures and include the appropriate units.

w =

Part C

What is the ratio of translational to rotational kinetic energy at the bottom?

Express your answer using three significant figures.

KEtr/KErot =

Part D

Avoid putting in numbers until the end so you can answer: do your answers in previous parts depend on the radius of the sphere or its mass? Select the best answer

a)

Only the angular speed depends on the radius. None of the results depend on the mass.

b) The angular and the translational speeds depend on the radius. None of the results depend on the mass.

c) Only the translational speed depends on the radius. None of the results depend on the mass.

d) The angular and the translational speeds depend on the radius and on the mass.

Answers

A:

mgh=\frac {1}{2}I\omega^2 +\frac {1}{2}mv^2

mgh=\frac {1}{2}\frac {2}{5}mr^2\left (\frac {v}{r} \right )^2 +\frac {1}{2}mv^2

mgh=\frac {7}{10}mv^2

v=\sqrt {\frac {10gh}{7}}

v=\sqrt {\frac {10\cdot 9.8\cdot 10\sin 30^{\circ}}{7}}\: m/s

v=8.37\: m/s

B:
\omega =\frac {v}{r}=\frac {8.37}{0.345}\: rad/s= 24.26\: rad/s

C:

K_r=\frac {1}{5}mv^2

K_t=\frac {1}{2}mv^2

\frac {K_t}{K_r}=\frac {5}{2}

D:

(a)

Only the angular speed depends on the radius. None of the results depends on the mass.


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