Can someone check my answers to these pretty simple (to you guys)
homework problems. They're due tomorrow, but I just want to make sure
they're correct before submitting them.

Question 1
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A barrel is placed at the top of a steep hill. The hill is 200m long
(as measured along the ground) and makes an angle of 30deg with the
horizontal. How fast will the barrel be travelling as it reaches the
end of the hill if it is released from rest? How long will it take for
the barrel to reach the bottom of the hill?

My solution
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I treat the barrel as a perfect cylinder of mass M and radius R. Then
the moment of inertia of the barrel is (1/2)MR^2. I use conservation
of energy to solve the problem.

At the top of the hill the energy of the barrel is in the form of
potential energy:

E1 = Mgh = MgxSin(theta)

where x is the distance along the hill from the initial point, g is
the acceleration due to gravity, and theta=30 degrees.

At the bottom of the hill I choose the potential energy to be zero, so
the total energy of the barrel here is given by

E2 = (rotational kinetic energy) + (kinetic energy of center of mass)

which is equal to

E2 = (1/2)Iw^2 + (1/2)mv^2

where w=v/R is the angular velocity of the barrel about its center, I
is the moment of inertia, and v is its velocity down the hill. I can
rearrange E2 to get

E2 = (1/2)v^2(I/R^2 + m)

Conservation of energy then implies that E1 = E2, so rearranging for v
gives me

v = sqrt( (4/3)gxSin(theta) )
= 37 metres per second.

Is this correct? Finding the time taken for the barrel to reach the
bottom of the slope is then a trivial matter (I hope).

Thanks in advance to anyone who helps!