0.35 Phy1350: angular momentum -- the physics of rolling objects  (Page 4/5)

M1*g*h = K1 = (1+Q1)*M1*(V1cm)^2, and

M2*g*h = K2 = (1+Q2)*M2*(V2cm)^2

Eliminating the mass from both sides of both equations yields

g*h = (1+Q1)*(V1cm)^2, and

g*h = (1+Q2)*(V2cm)^2

Solving each equation for the velocity yields

V1cm = ((g*h)/(1+Q1))^(1/2), and

V2cm = ((g*h)/(1+Q2))^(1/2)

Taking the ratios of the velocities yields

V1cm/V2cm = ((1+Q2)/(1+Q1))^(1/2)

Solving for V1cm in terms of V2cm yields

V1cm = V2cm * ((1+Q2)/(1+Q1))^(1/2)

From Figure 1 ,

For the cylindrical shell, Q1 = 1 and

For the solid cylinder, Q2 = (1/2)

Substituting values yields

V1cm = V2cm * ((1+(1/2))/(1+1))^(1/2)

Solving with the Google calculator yields

V1cm = V2cm * 0.866

Therefore, the translational velocity of the cylindrical shell is less than the translational velocity the solid cylinder by a factor of 0.866. This isbecause more of the potential energy is transformed into rotational kinetic energy in the cylindrical shell, resulting in less translational kinetic energyand less translational velocity.

Note that this result is independent of the mass of the objects, the radii of the objects, the height of the incline, and the angle of the incline.

Part 2

What percentage of the total kinetic energy is translational kinetic energy for each of the cylinders in Part 1 ?

Solution:

From above ,

K = (1 + Q)*(1/2)*M*(Vcm)^2

We know from earlier modules that

Ktr = (1/2)*M*(Vcm)^2

Taking the ratio of Ktr to K gives us

Ktr/K = ((1/2)*M*(Vcm)^2)/((1 + Q)*(1/2)*M*(Vcm)^2)

Simplification yields

Ktr/K = 1/(1+Q), or

Ktr = K * (1/(1+Q))

For the cylindrical shell

Ktr = K * (1/(1+1)) = K * (1/2)

Half or 50 percent of the total kinetic energy is translational kinetic energy for the cylindrical shell.

For the solid cylinder

Ktr = K * (1/(1+Q)) = K * (1/(1+(1/2))) = K * 0.67

Sixty-seven percent of the total kinetic energy is translational kinetic energy for the solid cylinder.

Acceleration of a rolling cylinder

A solid cylinder is rolling down an incline without slipping. What is the translational acceleration of the cylinder? How does that acceleration comparewith the acceleration of the same object sliding down the incline?

Assume that

• The angle of incline is represented by U
• The value of Q for the cylinder is (1/2)
• The coefficient of friction is unknown
• The mass of the cylinder is unknown
• The radius of the cylinder is unknown

Solution:

The only forces acting on the cylinder are the force of gravity and an unknown frictional force parallel to the incline and pointing up the incline.The force of gravity does not produce a torque on the object acting about the center of mass. Therefore, the only torque is produced by the unknown force offriction.

Angular acceleration is commonly represented by the Greek letter alpha. However, your Braille display probably won't display the Greek letter alpha, sowe will let angular acceleration be represented by "An" and will let translational acceleration be represented by "Atr".

We know that the relationship between torque, angular acceleration, and moment of inertia is

Sum of torques = I*An

We only have one torque. Therefore,