These next questions have to do with these two rolls of paper towels. 10) We pla

Question

These next questions have to do with these two rolls of paper towels. 10) We place both of these rolls at the top of a ramp and let them go. Which one reaches the bottom of the ramp first? The large roll, on the left. The small roll, on the right. Because they are the same shape, it's a tie. Suppose we let each of these rolls fall, but held on to the loose ends so that towels unroll as they fall. What would happen to their accelerations as they fell and continued to unroll? Their accelerations would increase, up to the limit of g. increase, up to the limit of 2/3 g. decrease, down to the limit of 1/2 g. stay the same all the way down. Which equation would allow you to calculate the speed of either roll after falling through a distance h? The image to the left shows the forces acting on a disk rolling down a ramp. When we found the acceleration of such a disk, we had to put the reference point for torques at its center. The result? We threw away the weight of the disk and were left with two unknowns, the friction force and the acceleration. Why did we have to do that? Torques must be taken around the center of the object. We eliminate two forces that way. We only knew 1 (iota) for an axis through the center. That's the natural center of rotation, so we must use it. That image is from a website at the prestigious Massachusetts Institute of Technology (MIT). Despite that, what is wrong with it? The normal force, N, is mg cos theta and so should be smaller than mg, not larger The friction force, f, is pN and so should be smaller than both N and mg Both a and b are true.

Explanation / Answer

(10)

(10) As the shape is same for both rolls so they will take same time to roll down the slope

Because their initial height is smae and their final velocities would also independent of its mass and radius

(11) When they unroll its shape would remain same so Its accelration would remain same

(12) initial potential energy = rotational kinetic energy + translational kinetic energy

mgh = 1/2 Iw2 + 1/2 mv2

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