The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass of 8.50kg. When outstretched, they span 1.80m; when wrapped, they form a cylinder of radius 25.0cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.450 kg·m^2. If the skater's original angular speed is 0.450 rev/s, what is his final angular speed?
The total moment of inertia with arms outstretched is:
I1 = Ibody + Iarms = 0.450 + (1/12) m L^2 = 0.450 + (1/12)*8.5*(1.8)^2 = 2.75 kg m^2
The total moment of inertia with arms rapped aound the body is:
I2 = Ibody + Iarms = 0.450 + m R^2 =
0.45 + 0.53 = 0.98 kg m^2
Use the conservation of angular momentum equation
I1*w1 = I2*w2 to get the final anglualr speed w2. It should increrase by about a factor of 3.
1.10
drwls thanks so much. your solution is absolutely right haha =) i got a similar qns to that of Erin too. and managed to solve it thanks to u.
Well, it seems like the skater is quite the physics expert, doing calculations while spinning! Let's see if I can help with the final angular speed.
First, let's find the moment of inertia of the skater's hands and arms when they are outstretched. We can consider them as a slender rod rotating about an axis through its center. The moment of inertia of a slender rod is given by I = (1/3) * M * L^2, where M is the mass and L is the length. Plugging in the values, we get:
I_outstretched = (1/3) * 8.50 kg * (1.80 m)^2
Now, when the skater wraps his arms, they form a thin-walled hollow cylinder. The moment of inertia of a thin-walled hollow cylinder is given by I = (1/2) * M * R^2, where M is the mass and R is the radius. Plugging in the values, we get:
I_wrapped = (1/2) * 8.50 kg * (0.25 m)^2
Now, let's calculate the total moment of inertia of the skater by adding the moment of inertia of the remainder of his body (0.450 kg·m^2) to the moment of inertia of his wrapped arms:
I_total = I_outstretched + I_wrapped + 0.450 kg·m^2
Now, we can use the conservation of angular momentum to find the final angular speed. The angular momentum is given by L = I * ω, where ω is the angular speed. Since the moment of inertia is constant, the initial angular momentum is equal to the final angular momentum:
I_outstretched * ω_outstretched = I_total * ω_final
Now we can rearrange the equation to solve for ω_final:
ω_final = (I_outstretched * ω_outstretched) / I_total
Plugging in the values, we can calculate the final angular speed. I'll leave the actual calculation to you, but remember to convert the angular speed from rev/s to rad/s. Good luck!
To solve this problem, we need to use the principle of conservation of angular momentum. Angular momentum is conserved when no external torque acts on the system.
The initial angular momentum (L_i) of the skater can be calculated using the formula:
L_i = I_i * ω_i
Where:
L_i is the initial angular momentum
I_i is the initial moment of inertia
ω_i is the initial angular velocity
The final angular momentum (L_f) can be calculated as:
L_f = I_f * ω_f
Where:
L_f is the final angular momentum
I_f is the final moment of inertia
ω_f is the final angular velocity
Since angular momentum is conserved, L_i = L_f.
Now let's calculate L_i:
The initial moment of inertia (I_i) can be split into two parts: the hands and arms (I_1) and the remainder of the body (I_2).
I_i = I_1 + I_2
For the hands and arms, we can use the formula for the moment of inertia of a thin-walled hollow cylinder:
I_1 = M_1 * r^2
Where:
M_1 is the mass of the hands and arms
r is the radius of the cylinder formed when wrapped
Substituting the given values, we get:
I_1 = 8.50kg * (0.25m)^2
Next, we need to find the moment of inertia for the remainder of the body, I_2. We are given that I_2 is constant and equal to 0.450 kg·m^2.
Now we can calculate the initial angular momentum, L_i:
L_i = (I_1 + I_2) * ω_i
Substituting the values, we get:
L_i = (8.50kg * (0.25m)^2 + 0.450 kg·m^2) * 0.450 rev/s
Since L_i = L_f, we can now solve for the final angular velocity, ω_f:
L_f = (I_1 + I_2) * ω_f
Substituting the known values and solving for ω_f:
(8.50kg * (0.25m)^2 + 0.450 kg·m^2) * 0.450 rev/s = (8.50kg * (0.25m)^2 + 0.450 kg·m^2) * ω_f
Simplifying the equation, we get:
0.450 rev/s = ω_f
Therefore, the skater's final angular speed is 0.450 rev/s.