A gob of clay, mass 217 g, falls from rest a distance 50 cm before striking and sticking to the edge of a wheel free to rotate about a horizontal axis through its center (Fig. 10-38). The wheel can be approximated as a solid disk of mass 49 kg and radius 84 cm. What is the angular speed of the wheel with the gob of clay attached?
(Answer: 0.0327rad/s)
Answer is provided above. Please show all work on how to get the answer.
m = 0.217kg
h = 0.5m
M = 49kg
r = 0.84m
mvr = Iw
1/2mv^2 = mgh
v = sqrt[2gh]
v = sqrt[2 * 9.8m/s^2 * 0.5m]
v = 3.1m/s
w = mvr / I
w = mvr / 1/2Mr^2
w = 2msqrt[2gh] / Mr
w = 2(0.217kg)(3.1m/s) / (49kg * 0.84m)
w = 0.033rad/s
To find the angular speed of the wheel with the gob of clay attached, we can use the principle of conservation of angular momentum.
The angular momentum is defined as the product of the moment of inertia and the angular velocity.
First, let's calculate the moment of inertia of the wheel. The moment of inertia (I) of a solid disk rotating about an axis through its center is given by the formula:
I = (1/2) * m * r^2
where m is the mass of the disk and r is the radius of the disk.
Given that the mass of the wheel is 49 kg and the radius is 84 cm (converting to meters, 0.84 m), we can substitute these values into the formula:
I = (1/2) * 49 kg * (0.84 m)^2
I ≈ 17.496 kg*m^2
Next, let's consider the angular momentum of the gob of clay. The initial angular momentum (L1) of the gob of clay is zero since it is at rest. After attaching to the wheel, the final angular momentum (L2) of the clay and wheel system will be the sum of their individual angular momenta.
L2 = I * ω
where ω is the angular velocity of the wheel.
Since angular momentum is conserved, we can equate the initial and final angular momenta:
L1 = L2
0 = I * ω
Solving for ω:
ω = 0 / I
ω = 0
This means that the wheel will not rotate initially due to the conservation of angular momentum.
However, when the gob of clay sticks to the edge of the wheel, it exerts a torque on the wheel which causes it to start rotating. The torque applied to the wheel is given by the equation:
τ = r * F
where r is the radius of the wheel and F is the force exerted by the clay.
The force exerted by the clay can be found using Newton's second law:
F = m * a
where m is the mass of the clay and a is the acceleration due to gravity.
Given that the mass of the clay is 217 g (converting to kg, 0.217 kg) and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the force:
F = 0.217 kg * 9.8 m/s^2
F ≈ 2.1266 N
Now we can calculate the torque:
τ = 0.84 m * 2.1266 N
τ ≈ 1.7874 N*m
Since torque is equal to the moment of inertia multiplied by the angular acceleration (τ = I * α), we can find the angular acceleration:
α = τ / I
α = 1.7874 N*m / 17.496 kg*m^2
α ≈ 0.102 rad/s^2
Finally, we can find the angular velocity (ω) using the equation:
ω^2 = ω0^2 + 2αθ
where ω0 is the initial angular velocity, α is the angular acceleration, and θ is the angular displacement (which is the angle covered by the clay on the wheel). Since both the initial angular velocity and the angular displacement are zero in this case, the equation simplifies to:
ω^2 = 2αθ
ω^2 = 2 * 0.102 rad/s^2 * 0 rad
ω^2 = 0
Solving for ω:
ω = √0
ω = 0 rad/s
So, the angular speed of the wheel with the gob of clay attached is initially zero, as the wheel does not rotate initially due to the conservation of angular momentum.