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

Answer:

$v_f = 4.22 m/s$

Explanation:

As we know by energy conservation

initial total energy = final total energy

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2 + mgL sin\theta$

so we have

$v = R \omega$

$I = mR^2$

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

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

$30.25 - 12.43 = v_f^2$

$v_f = 4.22 m/s$