two skaters, each of mass 55 kg, approach each other along parallel paths separated by 3.8 m. They have opposite velocities of 1.9 m/s each. One skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible.
(a) What is the radius of the circle they now skate in?
(b) What is the angular speed of the skaters?
(c) What is the kinetic energy of the two-skater system?
(d) Next, the skaters pull along the pole until they are separated by 0.7 m. What is their angular speed then?
(e) Calculate the kinetic energy of the system now.
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(a) The radius of the circle they skate in can be determined by using the conservation of angular momentum. Since the skaters are initially at rest vertically, the total angular momentum before and after the collision remains the same. We can calculate the initial angular momentum (L) as:
L = m1v1r1 + m2v2r2
Where m1 and m2 are the masses of the skaters, v1 and v2 are their velocities, and r1 and r2 are the distances from the center of the circle.
Since they have opposite velocities of 1.9 m/s, we can substitute these values into the equation:
L = (55 kg)(1.9 m/s)(r1) + (55 kg)(-1.9 m/s)(r2)
Since the radius of the circle they skate in is the same for both skaters, we can set r1 = r2 = R, where R is the radius we want to find:
L = (55 kg)(1.9 m/s)(R) + (55 kg)(-1.9 m/s)(R)
Simplifying the equation gives:
0 = 2(55 kg)(1.9 m/s)(R)
Therefore, R = 0 m. This means that the skaters do not form a circle and the skater carrying the pole does not rotate around the other skater.
(b) Since the skaters do not form a circle, they do not have an angular speed.
(c) Since the skaters do not form a circle, they do not have any rotational kinetic energy. However, they still have translational kinetic energy. The kinetic energy of the two-skater system can be calculated using the formula:
KE = 0.5(m1v1^2 + m2v2^2)
Substituting the values gives:
KE = 0.5(55 kg)(1.9 m/s)^2 + 0.5(55 kg)(-1.9 m/s)^2
Simplifying the equation gives:
KE = 0.5(55 kg)(3.61 m^2/s^2)
Therefore, the kinetic energy of the two-skater system is 1091 J.
(d) When the skaters pull along the pole until they are separated by 0.7 m, the radius of the circle they skate in changes. We can use the conservation of angular momentum again to determine their new angular speed.
Using the same equation as before:
L = m1v1r1 + m2v2r2
Substituting the values gives:
L = (55 kg)(1.9 m/s)(0.7 m) + (55 kg)(-1.9 m/s)(0.7 m)
Simplifying the equation gives:
L = 0
Since the angular momentum is zero, the skaters do not have any angular speed.
(e) Since the skaters do not have any angular speed, they do not have any rotational kinetic energy. However, they still have translational kinetic energy. Using the same formula as before:
KE = 0.5(m1v1^2 + m2v2^2)
Substituting the values gives:
KE = 0.5(55 kg)(1.9 m/s)^2 + 0.5(55 kg)(-1.9 m/s)^2)
Simplifying the equation gives:
KE = 0.5(55 kg)(3.61 m^2/s^2)
Therefore, the kinetic energy of the system now is still 1091 J.
To solve this problem, we can use the conservation of angular momentum. The initial angular momentum of the system remains constant throughout the entire motion since there are no external torques acting on it.
Let's break down the problem step by step:
(a) What is the radius of the circle they now skate in?
The initial situation involves the skaters rotating around a center of the pole. Since the skaters have opposite velocities and they grab onto the pole, the system's center of mass remains stationary. Now, the skaters rotate around the center of mass of the system.
To find the radius of the circle, we can use the formula for the angular momentum of the system:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Since the skaters are rotating around the center of mass of the system, the moment of inertia can be approximated as the sum of the individual moments of inertia of the skaters:
I = m1 * r1^2 + m2 * r2^2
where m1 and m2 are the masses of the skaters, and r1 and r2 are the distances from the skaters to the center of mass.
Given that the skaters have opposite velocities, their distance from the center of mass is half of the separation between them:
r1 = r2 = 3.8 m / 2 = 1.9 m
Plugging in the values into the equation for the moment of inertia:
I = (55 kg * (1.9 m)^2) + (55 kg * (1.9 m)^2)
I = 2 * (55 kg) * (1.9 m)^2
I = 470.6 kg*m^2
Since the angular momentum is conserved, we can equate the initial angular momentum (before grabbing the pole) to the final angular momentum (after grabbing the pole):
L_initial = L_final
m1 * r1 * v1_initial + m2 * r2 * v2_initial = I * ω_final
Since the skaters have opposite velocities, v1_initial = -v2_initial = -1.9 m/s:
(55 kg * 1.9 m) * (-1.9 m/s) + (55 kg * 1.9 m) * (1.9 m/s) = 470.6 kg*m^2 * ω_final
Simplifying:
(-1.9 m/s + 1.9 m/s) * (55 kg * 1.9 m) = 470.6 kg*m^2 * ω_final
0 m/s * (55 kg * 1.9 m) = 470.6 kg*m^2 * ω_final
0 = 470.6 kg*m^2 * ω_final
Therefore, the angular speed of the skaters is zero, which means they are not rotating around a circle.
(b) What is the angular speed of the skaters?
As calculated above, the angular speed of the skaters is zero since they are not rotating around a circle.
(c) What is the kinetic energy of the two-skater system?
The kinetic energy of the system can be calculated by summing the kinetic energies of the individual skaters:
KE = KE1 + KE2
where KE1 and KE2 are the kinetic energies of the first and second skater, respectively.
The kinetic energy of each skater can be calculated using the formula:
KE = 0.5 * m * v^2
where m is the mass of the skater and v is their velocity.
Substituting the given values:
KE = 0.5 * 55 kg * (1.9 m/s)^2 + 0.5 * 55 kg * (1.9 m/s)^2
Simplifying:
KE = 0.5 * 55 kg * (1.9 m/s)^2 * (1 + 1)
KE = 0.5 * 55 kg * (1.9 m/s)^2 * 2
KE = 90.55 J
Therefore, the kinetic energy of the two-skater system is 90.55 Joules.
(d) Next, the skaters pull along the pole until they are separated by 0.7 m. What is their angular speed then?
When the skaters pull along the pole, the conservation of angular momentum still applies. However, the moment of inertia of the system changes since the separation between skaters decreases.
To find the new angular speed, we can repeat the same steps as in part (a) but with the new separation distance of 0.7 m.
Using the same formula for the moment of inertia:
I = m1 * r1^2 + m2 * r2^2
we can substitute the new separation distance:
r1 = r2 = 0.7 m / 2 = 0.35 m
Continuing with the calculations:
I = (55 kg * (0.35 m)^2) + (55 kg * (0.35 m)^2)
I = 2 * (55 kg) * (0.35 m)^2
I = 20.075 kg*m^2
Now, we can use the conservation of angular momentum:
m1 * r1 * v1_initial + m2 * r2 * v2_initial = I * ω_final
Substituting the given values:
(55 kg * 0.35 m) * (-1.9 m/s) + (55 kg * 0.35 m) * (1.9 m/s) = 20.075 kg*m^2 * ω_final
Simplifying:
(-0.93425) + (0.93425) = 20.075 kg*m^2 * ω_final
0 = 20.075 kg*m^2 * ω_final
Therefore, the angular speed of the skaters is also zero when they are separated by 0.7 m.
(e) Calculate the kinetic energy of the system now.
Since the skaters have zero angular speed, they are no longer rotating and their kinetic energy remains the same as calculated in part (c).
Therefore, the kinetic energy of the system is still 90.55 Joules.
(a) 1.9 m
(b) w = V/R = 1.0 rad/s
(c) 55*(1.9)^2 Joules
(d) angular momentum is conserved and they speed up by a factor 3.8/0.7
(e) Do the numbers. Kinetic energy increases by a factor equal to the square of the velocity incorease factor.