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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?

Respuesta :

Answer:

Explanation:

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

onyebuchinnaji

The relationship between the angular speed and linear velocity of the cylinder is v = ωr.

The total kinetic energy of the cylinder at any velocity, v, is ³/₄mv².

Relationship between the angular speed and linear velocity

The relationship between the angular speed and linear velocity of the cylinder is given as;

v = ωr

where;

  • v is the linear velocity of the center mass
  • ω is rotational speed or angular speed

Total kinetic energy of the cylinder

The total kinetic energy of the cylinder is the sum of the kinetic energy and rotational kinetic energy of the cylinder.

K.E(total) = K.E(v) + K.E(rot)

K.E(total) =  ¹/₂mv² + ¹/₂Iω²

where;

  • I is the moment of the cylinder = ¹/₂mr²
  • ω is angular speed = v/r

K.E(total) =  ¹/₂mv² + ¹/₂(¹/₂mr²)(v/r)²

K.E(total) = ¹/₂mv² + ¹/₄mv²

K.E(total) = ³/₄mv²

Learn more about kinetic energy here: https://brainly.com/question/25959744

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