Translational Speed of Cylinder on Inclined Plane: Find the Answer

[bns1201]

Homework Statement

A solid cylinder with a mass of 1.12 kg and a radius of 0.017 m starts from rest at a height
of 3.40 m and rolls down a 77.7◦ slope. What is the translational speed of the cylin-
der when it leaves the incline? The acceleration of gravity is 9.81 m/s2 . Answer in units
of m/s.

Homework Equations

I have angular speed which is
v=r$$\omega$$

we know r (.017m). But I'm having trouble finding the others.
Any help would be great. Thanks.

 

[method_man]

You need to apply the work-energy principle:
P.E. lost by cylinder=K.E. gained by cylinder
P.E. lost=mgh
K.E. gained= 1/2*mv²+1/2*Iω²
v=rω, I=1/2*mr²

You have got the same problem https://www.physicsforums.com/showthread.php?t=147012".

 

[nlightner]

I would approach this problem by first identifying the relevant physical principles and equations that apply to the situation. In this case, we can use the principles of rotational motion and energy conservation.

First, we need to determine the moment of inertia of the cylinder, which is a measure of its resistance to rotational motion. For a solid cylinder, the moment of inertia is given by I = 1/2 * m * r^2, where m is the mass of the cylinder and r is its radius. Plugging in the given values, we get I = 1/2 * 1.12 kg * (0.017 m)^2 = 1.42*10^-4 kg*m^2.

Next, we can use the principle of conservation of energy, which states that the total energy of a system remains constant. At the top of the incline, the cylinder has potential energy given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height of the incline. At the bottom of the incline, the cylinder has both translational and rotational kinetic energy given by 1/2 * mv^2 and 1/2 * I * ω^2, respectively. Since the cylinder is rolling without slipping, we can relate the linear speed v and angular speed ω using the equation v = r*ω.

Setting the initial potential energy equal to the final kinetic energy, we can solve for the translational speed of the cylinder:

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

Substituting in the given values, we get:

1.12 kg * 9.81 m/s^2 * 3.40 m = 1/2 * 1.12 kg * v^2 + 1/2 * 1.42*10^-4 kg*m^2 * (v/0.017 m)^2

Solving for v, we get v = 3.37 m/s.

Therefore, the translational speed of the cylinder when it leaves the incline is 3.37 m/s.