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Rotational and Linear Speed & Kinetic Energy

A full keg of beer weighs 170 lbf. The beer inside the keg weighs 124lbf and the container weighs 46lbf. The keg is placed at the top of ramp length L=8ft. from the back of your pickup truck to the ground and released.

1. The rotational and linear speed of the keg at the bottom of the ramp.
2. The kinetic engery of the keg at the bottom of the ramp.
3. The time it takes the keg to get to the bottom of the ramp.
4. The magnitude of the friction force between the keg and the ramp.


1. The coefficient of friction between the ramp and the keg is 0.35.
2. The height of the bed of the pickup truck is H=20 inches.
3. The container can be modeled as a perfect cylinder with a radius of 9 inches and no end caps.
4. The container rotates, but the beer inside does not.


1. The friction force is LESS than the coefficient of friction times the normal force.

Solution Preview

Please see the attached file.


Ramp length 8ft

Mgsinθ 20 inches = 1.67ft

θ Mg

Let the mass of the keg be m1 and that of the beer be m2. Total mass M = m1+ m2.

Various forces acting on the beer keg are shown in the fig.. Total weight Mg is resolved into components Mgcosθ perpendicular to the ramp and Mgsinθ parallel to the ramp. Normal reaction N = Mgcosθ. F is the frictional force on the keg surface from the ramp.

Net linear force acting on the keg parallel to the ramp = Mgsinθ -F. Under this net force the keg starts to accelerate linearly down the ramp. In other words, this net force provides the keg with linear acceleration (and velocity) parallel to the ramp.

Linear acceleration of the keg down the ramp = a = (Mgsinθ ...

Solution Summary

The problem has been solved step by step explaining the fundamentals at each step. Rational and linear speed/kinetic energy is analyzed.