Question

A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ slope. What is the translational speed of the cylinder when it leaves the incline? The acceleration of gravity is 9.81 m/s 2 .

Answer 1

`v_(f)\approx 2.097\,(m)/(s)`

Step-by-step explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

`m_(cyl)\cdot g\cdot y_(o) = (1)/(2)\cdot m_(cyl) \left( 1+ (1)/(R)\right)\cdot v_(f)^(2)`

`g\cdot y_(o) = (1)/(2)\cdot\left( 1+ (1)/(R)\right)\cdot v_(f)^(2)`

The translational speed is:

`v_(f) = \sqrt{(2\cdot g\cdot y_(o))/(\left(1 + (1)/(R) \right))}`

`v_(f) = \sqrt{(2\cdot (9.807\,(m)/(s^(2)) )\cdot (4.80\,m))/(\left(1 + (1)/(0.049\,m) \right)) }`

`v_(f)\approx 2.097\,(m)/(s)`