A spherical pinball, initially at rest, rolls down a ramp inclined at 18 degrees

Question

A spherical pinball, initially at rest, rolls down a ramp inclined at 18 degrees relative to horizontal.
A. What is its translational speed after it has traveled 15 cm down the ramp?
B. Determine the translational speed after 3 seconds. (This is tricky because you must express the distance traveled down the ramp and velocity using constant acceleration equations. Plug those into your equation relating the distance and velocity.)

Please be explicit in steps.. (I"m getting something different than my professor)

Explanation / Answer

Part A)

We need the height at 15 cm down the ramp

h = (15)(sin18) = 4.64 cm

Now apply conservation of energy

mgh = .5mv^2 + .5Iw^2

I for a sphere is 2/5mr^2 and w = v/r, thus .5Iw^2 becomes

.5(2/5mr^2)(v^2/r^2) = .2mv^2

Now mgh = .5mv^2 + .2mv^2 (mass cancels)

(9.8)(.0464) = .7v^2

v = .806 m/s

Part B)

From above, we can find the acceleration

vf^2 = vo^2 + 2ad

.806^2 = 0^2 + 2(a)(.15)

a = 2.165 m/s^2

Now we can find the distance after three seconds

d = vot + .5at^2

d = (0) + (.5)(2.165)(3^2)

d = 9.744 m

The height = 9.744(sin 18) = 3.01 m

Now apply the same as above

(9.8)(3.01) = .7v^2

v = 6.49 m/s

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