A ring of mass 20kg and radius 0.4 meters [ I=MR^2 ] starts from rest at a heigh

Question

A ring of mass 20kg and radius 0.4 meters [ I=MR^2 ] starts from rest at a height of 4.00 meters and rolls down a 30.0 degrees slope.

A) What is total mechanical energy )potential energy + rotational kinetic energy + transnational kinetic eergy) of the ring when it is at the top of the plane?

B) Write the expression for the total energy of the ring when it is at the bottom of the plane in terms of its linear velocity 'v'. Where 'v' is the linear velocity at the bottom of the plane.

C) Find the magnitude of the velocity of the ring when it is at the bottom of the plane.

E) If a circular object of the same mass and radiu [ I=(2/7)MR^2) ] as the above ring released from the same height. In a race between the anove two objects on the above incline plane, (a) which one would win (hint: you don't have to do calculations all over again to find the asnwer to this question). (b) Explain briefly why (you don't have to show calculations for this part)?

Explanation / Answer

A) Ei = m g h = 20*9.81*4=784.8 J

b)remember w = v/r
E = 1/2 mv^2 + 1/2 I w^2 =0.5*20*v^2 + 0.5*20*v^2 = 20 v^2

C) 784.8 = 20 v^2
v= sqrt(784.8/20)=6.26 m/s

E) a smaller I means less energy goes into rotation so it can go faster so the circular object will win

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