A solid tin can (solid cylinder) with a mass of 6.2 kg and a radius of 0.11 m is

Question

A solid tin can (solid cylinder) with a mass of 6.2  kg and a radius of 0.11  m is on the top of a hill. The hill has a height of 28  m . The solid tin can (solid cylinder) is initially at rest at the top of the hill and then rolls without slipping down to the bottom of the hill.

1. Which of the following is the correct formual for the moment of intertia for a solid tin can (solid cylinder)?

1/2MR2, 2/3MR2, MR2, 1/3MR2, or 2/5MR2

2. What is the moment of intertia for the solid tin can (solid cylinder)?

3. What is the velocity of the solid tin can (solid cylinder) at the bottom of the hill?

4. If the solid tin can (solid cylinder) does not roll, but instead slips so that it slides down the hill, what is the velocity at the bottom of the hill?

Explanation / Answer

1.) moment of inertia of solid cyliner = 0.5 MR^2

2.)MOI of solid tin = 0.5*6.2*0.11^2 = 0.03751 Kgm^2

3.) Applying conservation of energy

mgh = 0.5 I w^2 + 0.5 mV^2

Mgh = 0.5*0.5*M*R^2*V^2/R^2 + 0.5 MV^2

gh = 0.25*V^2 + 0.5V^2

9.81*28 = 0.25*V^2 + 0.5v^2

V = 19.13 m/s

4.)

mgh = 0.5mv^2

9.81*28 = 0.5V^2

V = 23.43 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.