A hollow basketball, with radius 13.1 cm and mass 0.260 kg, is released from rest at the top of a 3.60 m long ramp inclined at 26.0° above the horizontal. What is the ball's velocity when it reaches the bottom of the ramp?
figure out how far it fell
like
h = 3.6 sin 26.0
then the potential energy at the top = m g h
That will be the kinetic energy at the bottom
m g h = (1/2)m v^2 + (1/2) I omega^2
but v = omega*R
so
m g h = (1/2) m v^2 + (1/2) I (v^2/R^2)
I is the moment of inertial of a thin walled sphere of radius 0.131 and mass 0.260 about an axis through the center.
ah, note, you do not really need to know the mass :)
because I =(2/5) m R^2
m g h = (1/2) m v^2 + (1/2) (2/5)m R^2 (v^2/R^2)
g h = (1/2) v^2 + (1/5) v^2 = 0.7 v^2
v^2 =( g* 3.6 sin 26 )/ 0.7
To find the ball's velocity when it reaches the bottom of the ramp, we can use the principle of conservation of mechanical energy.
Step 1: Calculate the potential energy at the top of the ramp.
The potential energy (PE) of an object at a certain height is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height. In this case, the height is the vertical distance from the top of the ramp to the bottom, which is the same as the vertical displacement on the ramp (h = 3.60 m * sin(26.0°)). So, the potential energy at the top is PE = (0.260 kg) * (9.8 m/s^2) * (3.60 m * sin(26.0°)).
Step 2: Calculate the kinetic energy at the bottom of the ramp.
The kinetic energy (KE) of an object is given by the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity. At the bottom of the ramp, all of the potential energy is converted to kinetic energy (assuming no energy losses due to friction or air resistance), so we can equate the potential energy to the kinetic energy: PE = KE.
Step 3: Solve for the velocity.
Setting the potential energy equal to the kinetic energy, we have (0.260 kg) * (9.8 m/s^2) * (3.60 m * sin(26.0°)) = (1/2) * (0.260 kg) * v^2. We can rearrange this equation to solve for the velocity (v).
Firstly, simplify the equation: (0.260 kg) * (9.8 m/s^2) * (3.60 m * sin(26.0°)) = (0.130 kg) * v^2.
Next, divide both sides of the equation by (0.130 kg) to isolate v^2: [(0.260 kg) * (9.8 m/s^2) * (3.60 m * sin(26.0°))] / (0.130 kg) = v^2.
Finally, take the square root of both sides to solve for v: v = sqrt{[(0.260 kg) * (9.8 m/s^2) * (3.60 m * sin(26.0°))] / (0.130 kg)}.
Evaluating this expression will give the velocity (v) of the ball when it reaches the bottom of the ramp.