Question

A hoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 25° below the horizontal. What is the rotational kinetic energy of the hoop after it has rolled 3.0 m down as measured along the surface of the ramp?

Answer 1

Answer:

$K E = 44.98\ J$

Explanation:

given,

mass of hoop = 2.75 Kg

horizontal speed = 4.5 m/s

angle with horizontal = 25°

m g l sin θ = m ( v² - u² )

g l sin θ = ( v² - u² )

9.8 x 3 x sin 25° = ( v² - 4.5² )

12.425 = v² - 20.25

v² = 32.675

v = 5.72 m/s

Rotational

$K E = \dfrac {1}{2}I\omega^2$

$K E = \dfrac {1}{2}\ mR^2\omega^2$

$K E = \dfrac {1}{2}\ mv^2$

$K E = \dfrac {1}{2}\ 2.75 \times 5.72^2$

$K E = 44.98\ J$