A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp,inclined 15°,and rolls to the bottom without slipping.(For a cylinder I =
MR². ) The upper end of the ramp is 1.5 m higher than the lower end.Find the linear speed of the cylinder when it reaches the bottom of the ramp.(g = 9.8 m/s²)

Question

A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp,inclined 15°,and rolls to the bottom without slipping.(For a cylinder I =

MR². ) The upper end of the ramp is 1.5 m higher than the lower end.Find the linear speed of the cylinder when it reaches the bottom of the ramp.(g = 9.8 m/s²)

Quizplus · Accepted Answer

Using conservation of energy, we can write: 
mgh = (1/2)mv^2 + (1/2)Iω^2 
where m is the mass of the cylinder, h is the height of the ramp, v is the linear speed of the cylinder at the bottom of the ramp, I is the moment of inertia of the cylinder, and ω is the angular speed of the cylinder at the bottom of the ramp.

Since the cylinder rolls without slipping, we know that v = ωr, where r is the radius of the cylinder. We also know that for a solid cylinder, I = (1/2)mr^2. Substituting these into the above equation and simplifying, we get:

mgh = (1/2)mv^2 + (1/4)mv^2 
Solving for v, we get: 
v = sqrt(2gh/5)

Plugging in the given values, we get: 
v = sqrt(2*9.8*1.5/5) = 4.4 m/s

So the correct answer is (B).

A Solid Cylinder of Mass 3