Question

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the hoop after it has rolled 3.00 m up as measured along the surface of the ramp

Answer 1

`v_f = 4.22 m/s`

Step-by-step explanation:

As we know by energy conservation

initial total energy = final total energy

so we have

`(1)/(2)mv^2 + (1)/(2)I\omega^2 = (1)/(2)mv_f^2 + (1)/(2)I\omega_f^2 + mgL sin\theta`

so we have

`v = R \omega`

`I = mR^2`

`mv_i^2 = mv_f^2 + mgL sin\theta`

`5.50^2 = v_f^2 + (9.81)(3) sin25`

`30.25 - 12.43 = v_f^2`

`v_f = 4.22 m/s`