A piece of putty is initially located at point A on the rim of a grinding wheel rotating about a horizontal axis. The putty is dislodged from point A when the diameter through A is horizontal. The putty then rises vertically and returns to A the instant the wheel completes one revolution
Find the speed of a point on the rim of the wheel in terms of the acceleration due to gravity and the radius R of the wheel. (Use m, and g as necessary.) If the mass of the putty is m, what is the magnitude of the force that held it to the wheel? (Use m, and g as necessary.)
To find the speed of a point on the rim of the wheel, we can use the principle of conservation of energy.
When the putty is dislodged from point A, it rises vertically and returns to A. The potential energy of the putty at its highest point is equal to the potential energy at point A, and the kinetic energy at its highest point is equal to the kinetic energy at point A.
1. Potential energy at highest point:
At the highest point, the potential energy is given by:
PE = mgh
Where:
m = mass of the putty
g = acceleration due to gravity
h = height of the highest point, which is equal to 2R (since the diameter through A is horizontal)
So, PE = m * g * 2R
2. Kinetic energy at point A:
At point A, the kinetic energy is given by:
KE = (1/2)mv^2
Where:
v = velocity/speed of a point on the rim of the wheel
Since the putty returns to point A the instant the wheel completes one revolution, the distance traveled is equal to the circumference of the wheel.
Distance = 2πR
The time taken to complete one revolution is equal to the period of revolution of the wheel (T). So, the speed of a point on the rim of the wheel is:
v = Distance / Time
v = 2πR / T
The period of revolution (T) is the time taken for one complete revolution:
T = 2π/ω
Where:
ω = angular velocity of the wheel
The angular velocity (ω) can be found by relating it to the linear velocity (v) and the radius (R) of the wheel:
v = ωR
Substituting this into the equation for T:
T = 2π / (v / R)
T = 2πR / v
Now, substituting the value of T into the equation for v:
v = 2πR / (2πR / v)
v = v
The speed of a point on the rim of the wheel is equal to the speed itself.
3. Magnitude of the force that held the putty to the wheel:
The force that held the putty to the wheel is the centripetal force acting on the putty. This force is given by:
F = mv^2 / R
Substituting the value of v:
F = m(v^2) / R
F = m ((2πR) / T)^2 / R
Substituting the value of T:
F = m ((2πR) / (2πR / v))^2 / R
F = m (v^2) / R
The magnitude of the force that held the putty to the wheel is equal to m(v^2) / R.
Therefore, the speed of a point on the rim of the wheel is v, and the magnitude of the force that held the putty to the wheel is m(v^2) / R.