1. First consider a mass on an inclined slope of angle θ, and assume the motion is frictionless. Sketch this arrangement:
2. As the mass travels down the slope it travels a distance x parallel to the slope. The change in height of the mass is therefore xsinθ. By conserving energy, equate the change of gravitational potential energy, mgh = mgxsinθ, to the kinetic energy for the mass as it goes down the slope. Then rearrange this to find an expression for the velocity of the mass, v. Assume the mass begins at rest, so that v = 0 and that x = 0 when t = 0.
3. Now because v = dx dt , substitute this into the equation and form a differential equation that you can integrate: put all things with x on the left and constants and t on the right. Then integrate both sides, integrating x from 0 to d and t from 0 to T. Rearrange the final expression so that you have T as a function of constants and d. This tells you the time it would take the object to move down a slope of length d.
4. Now consider a cylinder that rolls down the hill (so instead of the surface being frictionless, we have enough friction to prevent any sliding). Here we must add an extra term to the kinetic energy to include the rotational kinetic energy. This is 1 2 Iω 2 , where I is the moment of inertia. (We haven’t covered rotations in this course yet, but you can just use the results given here.) We can use the fact that when rolling v = ωr for a circular shape with radius r. Add this term into the equations in (2) and write the new expression for energy below, and rearrange for v again.
5. The moment of inertia for a cylinder is 1 2mr2 . Put this into your equation and again solve for the T to roll a distance d as for the nonrolling case above. Is the time greater or less?
6. Consider rolling a solid cylinder (as above), a hollow pipe (moment of inertia of mr2 ) and a solid sphere (moment of inertia 2 5mr2 ), all of the same radius, down the slope at the same time. In what order would they reach the bottom?
Solution in the uploaded images
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