A cylinder with moment of inertia I about its center of mass, mass m, and radius r has a string wrapped around it which is tied to the ceiling (Figure 1). The cylinder's vertical position as a function of time is y (t). At time t=0 the cylinder is released from rest at a height h above the ground.

A) The string constrains the rotational and translational motion of the cylinder. What is the relationship between the angular rotation rate? and v, the velocity of the center of mass of the cylinder?

B) Remember that upward motion corresponds to positive linear velocity, and counterclockwise rotation corresponds to positive angular velocity.

C) Suppose that at a certain instant the velocity of the cylinder is v. What is its total kinetic energy, Ktotal, at that instant?

D) Suppose that at a certain instant the velocity of the cylinder is v. What is its total kinetic energy, Ktotal, at that instant?

Question

Physics

Timothy Morales

24 July, 15:54

A cylinder with moment of inertia I about its center of mass, mass m, and radius r has a string wrapped around it which is tied to the ceiling (Figure 1). The cylinder's vertical position as a function of time is y (t). At time t=0 the cylinder is released from rest at a height h above the ground.

A) The string constrains the rotational and translational motion of the cylinder. What is the relationship between the angular rotation rate? and v, the velocity of the center of mass of the cylinder?

B) Remember that upward motion corresponds to positive linear velocity, and counterclockwise rotation corresponds to positive angular velocity.

C) Suppose that at a certain instant the velocity of the cylinder is v. What is its total kinetic energy, Ktotal, at that instant?

D) Suppose that at a certain instant the velocity of the cylinder is v. What is its total kinetic energy, Ktotal, at that instant?

+4

Answers (1)

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Kelvin Blankenship · Answer 1

Let us study the downward movement of cylinder which accelerates as well as rotates.

A)

If v be the linear downward velocity of cm of cylinder and ω be angular velocity of cylinder

v = ωr, when there is no slippage of string around cylinder.

B &C)

Total kinetic energy = Rotational + linear

= 1/2 Iω² + 1/2 m v²

1/2 x1/2 mr²ω² + 1/2 m v²

= 1/4 mv² + 1/2 m v²

= 3/4 m v²

For downward acceleration,

mg - T = ma where T is tension in string.

Rotational movement

Torque = T x r

Tr = I α, I is moment of inertia and α is angular acceleration.

= I a/r

T = I a / r², Putting this value of T in earlier equation

mg - I a / r² = ma

a (I / r² + m) = mg

a = mg / (I / r² + m)

For cylinders

I =.5 mr²

a = g / (.5 + 1)

= g / 1.5