A sphere of radius r =34.5 cm and massm = 1.80 kg starts from rest and rolls wit

Question

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

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?

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?

The angular and the translational speeds depend on the radius. None of the results depend on the mass. The angular and the translational speeds depend on the radius and on the mass. Only the angular speed depends on the radius. None of the results depend on the mass. Only the translational speed depends on the radius. None of the results depend on the mass.

Explanation / Answer

A)

Here , let v be the linear speed of sphere

w is the angular speed of sphere ,

for rolling without slipping

v = r*w

Now, using conservation of energy

0.5 * mv^2 + 0.5 * I*w^2 = mg*l * sin(theta)

0.5 * mv^2 + 0.5 * 0.4 m*r^2 * v^2/r^2 = m*g*l*sin(theta)

0.7 * v^2 = 9.8*10 * sin(30)

v = 8.36 m/s^2

tha translational speed is 8.36 m/s^2

(B)

radius of sphere, r =34.5 cm

r = 0.345 m

we know,

v = r * w

w = 8.36/0.345

w = 24.23 rad/s

the rotational speed is 24.23 rad/s

(C)

ratio of translatinal energy to the rotational energy,E = 0.5 * m*v^2 / 0.5 * I * w^2

E = 0.5 * m*v^2 / (0.5 * 0.4 *m *r^2 * v^2/r^2)

E = 5/2

ratio of translatinal energy to the rotational energy is 5:2

(D)

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

as the equation for translational speed is independent of mass

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