-46 points HRW10 11 PO016 In the figure below, a cylindrical object of mass M an

Question

-46 points HRW10 11 PO016 In the figure below, a cylindrical object of mass M and radius R rolls smoothly from rest down a ramp and onto a horizontal section. From there it rolls off the ramp and onto the floor, landing a horizontal distance d 0.520 m from the end of the ramp. The initial height of the cylinder is H 0.950 m; the end of the ramp is at height h 0.10 m. The object consists of an outer cylindrical shell (of a certain uniform density) that is glued to a central cylinder (of a different uniform density). The rotational inertia of the object can be expressed in the al form! -mR2, but is not 0.5 as for a cylinder with a single uniform density. Determine .

Explanation / Answer

after leaving the ramp,

initial vertical velocity, v0y = 0

ay = - 9.81 m/s^2

y = yf - yi = 0 - h = - 0.10 m

Applying y = v0y t + ay t^2 /2 2

- 0.10 = - 9.81 t^2 /2

t = 0.143 sec

in horizontal,

v = d/t = 3.64 m/s

Now applying energy conservation,

PEi + KEi = PEf + KEf

M g H + 0 = M g h + (M v^2 /2 + I w^2 / 2)

2M g (H - h) = M v^2 + (beta M R^2)(v/R)^2

= M v^2 (1 + beta)

2 x 9.81 x (0.950 - 0.10) = 3.64^2 (1 + beta)

beta = 0.257 ........Ans

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