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 = 0.5MR². )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) A) 4.7 m/s B) 4.4 m/s C) 4.0 m/s D) 2.4 m/s

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem.

The initial potential energy of the cylinder is:

PEi = mgh = (3.0 kg)(9.8 m/s^2)(1.5 m) = 44.1 J

At the bottom of the ramp, all of the potential energy is converted to kinetic energy and rotational kinetic energy:

KEf + KEr = PEi

where KEf is the translational kinetic energy and KEr is the rotational kinetic energy.

The translational kinetic energy is given by:

KEf = (1/2)mv^2

where v is the linear speed of the cylinder.

The rotational kinetic energy is given by:

KEr = (1/2)Iω^2

where I is the moment of inertia of the cylinder about its axis (given by I = 0.5MR^2) and ω is the angular speed of the cylinder (related to the linear speed v by ω = v/R).

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

(1/2)mv^2 + (1/2)I(v/R)^2 = mgh

v^2 = (2/3)gh

v = √[(2/3)(9.8 m/s^2)(1.5 m)] ≈ 4.4 m/s

Therefore, the answer is B.

A Solid Cylinder of Mass 3