A small 0.13 g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 0.3 g and is 2.49 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 4.1 cm/s relative to the table. What is the angular speed of the bar just after the frisky insect leaps? The unit of angular speed is rad/s
To find the angular speed of the bar just after the bug leaps off, we can use the principle of conservation of angular momentum. The angular momentum before the bug jumps off will be equal to the angular momentum after the bug jumps off.
The formula for angular momentum is given by L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Let's calculate the angular momentum before the bug jumps off:
1. Convert the mass of the bug, bar, and length of the bar into kilograms:
- Mass of the bug, m_bug = 0.13 g = 0.13 * 10^(-3) kg = 0.00013 kg
- Mass of the bar, m_bar = 0.3 g = 0.3 * 10^(-3) kg = 0.0003 kg
2. Convert the length of the bar into meters: L_bar = 2.49 cm = 2.49 * 10^(-2) m = 0.0249 m
3. Moment of inertia for a thin uniform bar rotating about a pivot is given by I = (1/3) * m * L^2. Let's calculate the moment of inertia for the bar:
I_bar = (1/3) * m_bar * L_bar^2
4. Calculate the initial angular momentum:
L_initial = I_bar * ω_initial, where ω_initial is the initial angular speed of the bar.
Now we need to calculate the angular momentum after the bug jumps off. Since there is no external torque acting on the system, the total angular momentum is conserved.
After the bug jumps off, the only object rotating is the bar. Let's call the angular speed of the bar after the bug jumps off as ω_final.
5. Calculate the final angular momentum:
L_final = I_bar * ω_final
Since the angular momentum is conserved, we can set L_initial = L_final and solve for ω_final.
6. Set the initial and final angular momenta equal to each other:
I_bar * ω_initial = I_bar * ω_final
7. Cancel out I_bar from both sides of the equation:
ω_initial = ω_final
Therefore, the angular speed of the bar just after the bug leaps is equal to the initial angular speed, ω_initial.
To find ω_initial, we need to use the conservation of linear momentum.
The formula for linear momentum is given by p = m * v, where p is the linear momentum, m is the mass, and v is the linear speed.
8. Calculate the linear momentum of the bug:
p_bug = m_bug * v_bug, where v_bug is the linear speed of the bug relative to the table.
9. Calculate the linear momentum of the bar:
p_bar = m_bar * v_bar, where v_bar is the linear speed of the bar.
Since there is no external force or torque acting on the system, the total linear momentum is also conserved.
10. Set the initial and final linear momenta equal to each other:
p_bug + p_bar = 0
11. Solve for the linear speed of the bar:
v_bar = -p_bug / m_bar
Now we can use the linear speed of the bar to find ω_initial.
12. Calculate the initial angular speed using the formula ω_initial = v_bar / R, where R is the distance of the bug from the nail (equal to half the length of the bar).
R = L_bar / 2
13. Calculate ω_initial:
ω_initial = v_bar / R
Finally, we have the value of ω_initial, which is the angular speed of the bar just after the bug leaps off.
You can plug in the values given in the problem to calculate the final answer.
To find the angular speed of the bar just after the bug jumps off, we can apply the principles of conservation of angular momentum.
The angular momentum of the system (bug + bar) is conserved before and after the bug jumps off. Initially, the angular momentum is zero since both the bug and the bar are at rest.
To calculate the angular momentum, we consider the moment of inertia and angular velocity. The moment of inertia of the system is the sum of the moment of inertia of the bar and the moment of inertia of the bug.
The moment of inertia of a slender rod rotating about one end is given by the equation:
I = (1/3) * m * L²
Where:
I is the moment of inertia
m is the mass of the object
L is the length of the object
Substituting the given values:
m(bar) = 0.3 g = 0.3 * 10^-3 kg
L(bar) = 2.49 cm = 2.49 * 10^-2 m
m(bug) = 0.13 g = 0.13 * 10^-3 kg
L(bug) = 0 cm (since the bug jumps off)
The moment of inertia of the bar can be calculated as:
I(bar) = (1/3) * m(bar) * L(bar)²
The moment of inertia of the bug is zero since it jumps off and has no distance from the pivot.
Therefore, the total moment of inertia of the system is:
I(total) = I(bar) + I(bug) = I(bar) + 0 = I(bar)
Now, we can use the principle of conservation of angular momentum:
Before the bug jumps off:
L(initial) = 0
After the bug jumps off:
L(final) = I(total) * ω(final)
Since angular momentum is conserved,
L(initial) = L(final)
0 = I(bar) * ω(final)
Solving for ω(final):
ω(final) = 0 / I(bar) = 0
Therefore, the angular speed of the bar just after the bug jumps off is zero rad/s.