A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel and observes that drops of water fly off tangentially. She measures the heights reached by drops moving vertically (see figure). A drop that breaks loose from the tire on one turn rises vertically 54.0 cm above the tangent point. A drop that breaks loose on the next turn rises 51.0 cm above the tangent point. The radius of the wheel is 0.395 m.
(a) Why does the first drop rise higher than the second drop?
(b) Neglecting air friction and using only the observed heights and the radius of the wheel, find the wheel's angular acceleration (assuming it to be constant).
To understand why the first drop rises higher than the second drop, we need to consider the forces acting on the drops as they break loose from the tire and the motion they experience.
(a) When the bicycle wheel is turned upside down, the water drops are subject to two primary forces: gravity and centripetal force. Gravity acts vertically downward, pulling the drops towards the ground. The centripetal force acts inward towards the center of the wheel, keeping the drops in circular motion.
When a drop breaks loose from the tire, it inherits the rotational velocity of the wheel at that point. The velocity of the drops has two components: a radial component (towards the center of the wheel) and a tangential component (perpendicular to the radial direction).
Now, when the drops break loose, they continue to move in a circular path (due to the tangential velocity they possess) while being acted upon by gravity. As a result, the drops take a curved trajectory upwards.
The height to which a drop rises depends on the magnitude of its tangential velocity. Since the tangential velocity decreases with time as the wheel slows down, the drops that break loose later have less tangential velocity. As a result, the second drop rises to a lower height compared to the first drop.
(b) To find the wheel's angular acceleration, we can use the observed heights and the radius of the wheel. The height reached by the drops is directly related to their tangential velocity.
We can analyze the motion of the drops using the principle of conservation of mechanical energy. At the lowest point of their trajectory, the kinetic energy of the drops (due to tangential velocity) is converted into potential energy (due to the height reached).
Using the conservation of mechanical energy equation:
mgh = 0.5mv^2
where:
m = mass of the drop (which cancels out in the equation)
g = acceleration due to gravity (9.8 m/s^2)
h = height reached by the drop
v = tangential velocity of the drop
For the first drop:
h1 = 54.0 cm = 0.54 m
For the second drop:
h2 = 51.0 cm = 0.51 m
Combining these equations for the two drops, we have:
g * h1 = 0.5 * v1^2
g * h2 = 0.5 * v2^2
Dividing the two equations:
(h1 / h2) = (v1^2 / v2^2)
Substituting the values:
(0.54 / 0.51) = (v1^2 / v2^2)
Rearranging the equation:
v1^2 = (0.54 / 0.51) * v2^2
The tangential velocity is related to the angular velocity (ω) by the equation v = rω, where r is the radius of the wheel.
Substituting v = rω in the equation and rearranging, we have:
(r^2 * ω1^2) = (0.54 / 0.51) * (r^2 * ω2^2)
As the radius of the wheel (r) is common to both sides of the equation, it cancels out:
ω1^2 = (0.54 / 0.51) * ω2^2
Finally, as angular acceleration (α) is the time derivative of angular velocity (ω), we can write:
α = (ω1 - ω2) / t
where t is the time taken for one complete revolution of the wheel.
Using the given information, we can calculate the angular acceleration by substituting the respective values for ω1, ω2, and t into the equation.