A 2.00 kg hoop with a radius of 0.0750 m was placed at the top of a 3.00 m long

Question

A 2.00 kg hoop with a radius of 0.0750 m was placed at the top of a 3.00 m long ramp. The top of the ramp was 0.800 m above the lower end of the ramp. The hoop rolled down the ramp.
a) Calculate the hoop’s total mechanical energy at the top of the ramp.
b) What is the hoop’s total mechanical energy at the bottom of the ramp?
c) Calculate the hoop’s speed at the bottom of the ramp.
d) Calculate the hoop’s angular velocity at the bottom of the ramp.
e) Calculate the hoop’s acceleration (linear acceleration).
f) Calculate the hoop’s angular acceleration.
g) Calculate the time required for the hoop to roll down the ramp.
h) Calculate the hoop’s kinetic energy (linear kinetic energy) at the bottom of the ramp.
i) Calculate the hoop’s rotational kinetic energy at the bottom of the ramp.

Explanation / Answer

(a)   the hoop’s total mechanical energy at the top of the ramp is E = mgh      = 2.0*9.8*0.8      = 15.68 J (b) According to conservation energy, the total mechanical energy remains constant.    the hoop’s total mechanical energy at the bottom of the ramp is 15.68 J. (c) Applying the conservation of energy, mgh = 1/2mv^2 + 1/2I^2          = 1/2mv^2 +1/2mv^2      (since = v/r)    gh   = v^2     v = 7.84 m/s (d) the angular velocity = v/r    = 7.84/0.075       = 104.53 rad/s (e) Applying the equations of motion, V^2 = 2a(3.0) a = 10.24 m/s^2 (f) the angular acceleration = a/r        = 10.24/0.075           = 136.59 rad/s^2 (g) From equations of motion, t^2 = 2(3.0)/a      t = 0.765 s (h) the K.E = 1/2mv^2              = 0.5*2*(7.84)^2             = 7.84 J (i) the rotational energy E(rotational) = 15.68 - 7.84                       = 7.84 J

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