Rolling Hoop Speed Calculation

What is the speed of the hoop after it has rolled 3.00 m up as measured along the surface of the ramp?

Given data: 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.

Answer: $v_f = 4.22 m/s

Answer:

By using the principle of energy conservation, we can determine the final speed of the hoop as it rolls up the ramp. The initial total energy must be equal to the final total energy.

Let's denote the initial speed as $v_i$, the final speed as $v_f$, the radius of the hoop as $R$, the moment of inertia as $I$, and the height of the ramp as $L$.

We have the equations:

$v = R \omega$

$I = mR^2$

$\frac{1}{2}m v_i^2 = \frac{1}{2}m v_f^2 + mgL \sin{\theta}$

Plugging in the given values, we can solve for the final speed $v_f$:

$5.50^2 = v_f^2 + (9.81)(3) \sin{25}$

$30.25 - 12.43 = v_f^2$

$v_f = 4.22 m/s$

When a hoop is rolling without slipping along a horizontal surface and then starts to move up a ramp at an angle, we can use the conservation of energy principle to determine its final speed. In this case, the initial kinetic energy of the hoop is equal to the final kinetic energy and potential energy when it reaches a certain height up the ramp.

The equation $\frac{1}{2}m v_i^2 = \frac{1}{2}m v_f^2 + mgL \sin{\theta}$ comes from the conservation of mechanical energy, where $m$ is the mass of the hoop, $v_i$ is the initial speed, $v_f$ is the final speed, $g$ is the acceleration due to gravity, $L$ is the height of the ramp, and $\theta$ is the angle of the ramp.

By solving the equation, we can determine that the final speed of the hoop after rolling 3.00 m up the ramp is $4.22 m/s$.

This calculation shows how energy principles can be applied to solve motion problems involving rolling objects on inclines.