Equate the increase in potential energy at the higher elevation, M g H, to the decrease in kinetic energy. Make sure you include the kinetic energy of rotation, which is (1/2) I w^2. For a sphere of radius R,
V = R w
and I =(2/5) I R^2
Therefore
KE(rotational) = (1/2)(2/5)M R^2(V/R)^2 = (1/5) M V^2
M g H = KE(translational) + KE(rotational)= (7/10)MV^2
V^2 = sqrt[(10/7)gH]
A bowling ball encounters a 0.76m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. If the translational speed of the ball is 3.80m/s at the bottom of the rise, find the translational speed at the top.
Where does the translational speed at the bottom come in?
A bicycle wheel has a diameter of 47.6 cm and a mass of 0.809 kg. The bicycle is placed on a stationary stand on rollers and a resistive force of 60.1 N is applied to the rim of the tire. Assume all the mass of the wheel is concentrated on the outside radius.
In order to give the wheel an acceleration of 3.18 rad/s2, what force must be applied by a chain passing over a sprocket with diameter 4.16 cm?
Answer in units of N.
In this problem, the equation for kinetic energy (KE) is mentioned, which takes into account both translational and rotational kinetic energy. The translational speed at the bottom of the rise is given as 3.80 m/s.
To find the translational speed at the top of the rise, we can use the principle of conservation of mechanical energy. According to this principle, the total mechanical energy (sum of potential energy and kinetic energy) remains constant in the absence of non-conservative forces like friction.
At the bottom of the rise, the ball has only translational kinetic energy and no potential energy. The equation for kinetic energy at the bottom is:
KE(bottom) = (7/10)MV^2,
where V is the translational speed at the bottom.
At the top of the rise, the ball has gained potential energy while losing translational kinetic energy due to the increase in height. The equation for potential energy at the top is:
PE(top) = Mgh,
where M is the mass of the ball, g is the acceleration due to gravity, and h is the vertical rise.
According to the principle of conservation of mechanical energy:
KE(bottom) + PE(bottom) = KE(top) + PE(top),
Since the ball starts from rest at the top, the initial kinetic energy at the top is zero, i.e., KE(top) = 0. Therefore:
(7/10)MV^2 + 0 = 0 + Mgh.
Simplifying the equation, we get:
(7/10)V^2 = gh.
Now, we can solve for V, which is the translational speed at the top of the rise:
V^2 = (10/7)gh,
V = sqrt[(10/7)gh].
Substituting the given values, where g is the acceleration due to gravity and h is the vertical rise of 0.76 m, you can calculate the translational speed at the top of the rise.