A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18° to the horizontal.How fast is it moving after it has rolled 2.2 m down the plane?
A)4.3 m/s
B)5.2 m/s
C)3.0 m/s
D)3.7 m/s
E)2.6 m/s

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. The initial potential energy of the cylinder is converted to kinetic energy as it rolls down the inclined plane. The equation for conservation of energy is:

PEi = KEf
mgh = 1/2mv^2 + 1/2Iω^2

where m is the mass of the cylinder, h is the height of the inclined plane, v is the velocity of the cylinder, I is the moment of inertia of the cylinder, and ω is the angular velocity of the cylinder.

Since the cylinder is rolling without slipping, we can use the relationship between linear and angular motion: v = rω, where r is the radius of the cylinder.

We can express the moment of inertia in terms of the mass and radius: I = 1/2mr^2, since the cylinder is uniform.

Substituting these relationships into the conservation of energy equation and solving for v, we get:

v = (2gh/7)^0.5

Plugging in the values given in the problem, we get:

v = (2*9.81*sin18*2.2/7)^0.5 = 3.0 m/s

Therefore, the answer is C.

A Uniform Solid 5