A 0.5 kg mass is moving in a circle of radius 1.8 m on a flat frictionless table at the end of a string. The speed of the mass is 2.6 m/s. The string routes through a hole in the center of the table and is held by you underneath the table.
What is the angular momentum of the mass?
What is the tension in the string?
If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string?
How much work did you do to reduce the radius by a factor of one-half?
To find the angular momentum of the mass, we use the formula:
Angular Momentum (L) = Moment of Inertia (I) * Angular Velocity (ω)
However, we need to find the Moment of Inertia, which is the resistance of a rotating object to changes in its rotation. For a point mass moving in a circle, the moment of inertia can be calculated as the product of the mass (m) and the square of the radius (r^2).
Moment of Inertia (I) = m * r^2
Plugging in the given values:
Mass (m) = 0.5 kg
Radius (r) = 1.8 m
I = 0.5 kg * (1.8 m)^2
I = 0.5 kg * 3.24 m^2
I = 1.62 kg·m^2
The angular velocity (ω) can be calculated by dividing the speed of the mass by its radius.
Angular Velocity (ω) = Speed (v) / Radius (r)
Plugging in the given values:
Speed (v) = 2.6 m/s
Radius (r) = 1.8 m
ω = 2.6 m/s / 1.8 m
ω ≈ 1.44 rad/s
Now we can calculate the angular momentum:
L = I * ω
L = 1.62 kg·m^2 * 1.44 rad/s
L ≈ 2.33 kg·m^2/s
To find the tension in the string, we use the concept of centripetal force. In circular motion, the centripetal force acts towards the center of the circle and is equal to the mass of the object multiplied by the square of its velocity, divided by the radius.
Centripetal Force (F) = (m * v^2) / r
Plugging in the given values:
Mass (m) = 0.5 kg
Speed (v) = 2.6 m/s
Radius (r) = 1.8 m
F = (0.5 kg * (2.6 m/s)^2) / 1.8 m
F ≈ 1.49 N
The tension in the string is equal to the centripetal force.
Now, let's consider the scenario when you pull on the string to decrease the radius to one-half its former value, which means the new radius (r_new) becomes 0.9 m.
To calculate the new tension in the string, we use the same formula for centripetal force, but with the new radius:
New Tension (T_new) = (m * v^2) / r_new
Plugging in the given values:
Mass (m) = 0.5 kg
Speed (v) = 2.6 m/s
New Radius (r_new) = 0.9 m
T_new = (0.5 kg * (2.6 m/s)^2) / 0.9 m
T_new ≈ 3.32 N
So, the new tension in the string is approximately 3.32 N.
To calculate the work done to reduce the radius by a factor of one-half, we can use the work-energy principle. The work done is equal to the change in kinetic energy.
The initial kinetic energy (K_initial) can be calculated using the formula:
K_initial = (1/2) * m * v^2
Plugging in the given values:
Mass (m) = 0.5 kg
Speed (v) = 2.6 m/s
K_initial = (1/2) * 0.5 kg * (2.6 m/s)^2
K_initial ≈ 1.69 J
The final kinetic energy (K_final) can be calculated using the new speed (v_new), which we can find using the conservation of angular momentum.
Angular Momentum (L) = Moment of Inertia (I) * Angular Velocity (ω)
Since the moment of inertia remains constant, the angular momentum should be conserved. Thus:
L_initial = L_final
I * ω_initial = I * ω_final
ω_initial = ω_final
Using this relationship, we can find the new angular velocity (ω_new):
ω_new = ω_initial = ω = 1.44 rad/s
The new speed (v_new) can be found by multiplying the new angular velocity (ω_new) by the new radius (r_new):
v_new = ω_new * r_new
v_new = 1.44 rad/s * 0.9 m
v_new ≈ 1.3 m/s
Now, we can calculate the final kinetic energy (K_final) using the formula:
K_final = (1/2) * m * v_new^2
Plugging in the given values:
Mass (m) = 0.5 kg
Speed (v_new) ≈ 1.3 m/s
K_final = (1/2) * 0.5 kg * (1.3 m/s)^2
K_final ≈ 0.845 J
The work done to reduce the radius is the change in kinetic energy:
Work = K_final - K_initial
Work ≈ 0.845 J - 1.69 J
Work ≈ -0.845 J
The negative sign indicates that work is done on the system (the mass) to reduce the radius.