A small 𝑚𝑏 = 09.0 ∗ 10−3 bug stands at one end of a thin uniform bar that is initially at rest on asmooth horizontal table. The other end of the bar pivots about a nail driven into the table and canrotate freely, without friction. The bar has a mass of 𝑀𝑏𝑎𝑟 = 49.0 ∗ 10−3𝑘𝑔 and is 𝐿𝑏𝑎𝑟 =100. 𝑐𝑚 in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with aspeed of 𝑣𝑏 = 21.0 𝑐𝑚𝑠 relative to the table. What is the angular speed of the bar just after the insectleaps? What is the total kinetic energy of the system just after the bug leaps? Where does thisenergy come from? (3 – sig figs)Problem 2The angular position 𝜃 of a 0.36m diameter flywheel is given by 𝜃 = (2.0 𝑟𝑎𝑑𝑠3 ) 𝑡3 . Find the 𝜃in radians and in degrees, at 𝑡1 = 2.0𝑠 and 𝑡2 = 5.0𝑠. (3 sig. figs)

Problem 1:

Step 1: We can use the conservation of angular momentum to solve this problem. The initial angular momentum of the system is zero because everything is at rest. The final angular momentum of the system is the sum of the angular momentum of the bug and the bar.

Step 2: The angular momentum of the bug is given by L_bug = m_bug * v_bug * L_bar. Substituting the given values, we get L_bug = 0.009 kg * 0.21 m/s * 1 m = 0.00189 kg*m^2/s.

Step 3: The angular momentum of the bar is given by L_bar = I_bar * w_bar, where I_bar is the moment of inertia of the bar and w_bar is the angular speed of the bar. The moment of inertia of the bar is given by I_bar = M_bar * L_bar^2 / 3. Substituting the given values, we get I_bar = 0.049 kg * (1 m)^2 / 3 = 0.01633 kg*m^2.

Step 4: Since the total angular momentum is conserved, we have L_bug = L_bar. Therefore, 0.00189 kgm^2/s = 0.01633 kgm^2 * w_bar. Solving for w_bar, we get w_bar = 0.00189 kgm^2/s / 0.01633 kgm^2 = 0.116 s^-1.

Step 5: The total kinetic energy of the system just after the bug leaps is the sum of the kinetic energy of the bug and the bar. The kinetic energy of the bug is given by K_bug = 1/2 * m_bug * v_bug^2 = 1/2 * 0.009 kg * (0.21 m/s)^2 = 0.00019845 J. The kinetic energy of the bar is given by K_bar = 1/2 * I_bar * w_bar^2 = 1/2 * 0.01633 kg*m^2 * (0.116 s^-1)^2 = 0.0001102 J. Therefore, the total kinetic energy is K_total = K_bug + K_bar = 0.00019845 J + 0.0001102 J = 0.00030865 J.

Step 6: The energy comes from the work done by the bug when it jumps off the bar. The bug applies a force over a distance, which does work on the system and increases its kinetic energy.

Problem 2:

Step 1: The angular position theta is given by the equation theta = (2.0 rad/s^3) * t^3.

Step 2: To find theta at

Problem 1:

Step 1: We need to use the conservation of angular momentum. The initial angular momentum is zero because the system is at rest. The final angular momentum is the sum of the angular momentum of the bug and the bar.

Step 2: The angular momentum of the bug is m_b * v_b * L_bar (mass times velocity times distance from the pivot point). The angular momentum of the bar is I_bar * w_bar (moment of inertia times angular speed).

Step 3: The moment of inertia of the bar is (1/3) * M_bar * L_bar^2.

Step 4: Setting the initial and final angular momentum equal to each other gives us: 0 = m_b * v_b * L_bar + (1/3) * M_bar * L_bar^2 * w_bar.

Step 5: Solving for w_bar (the angular speed of the bar) gives us: w_bar = - (m_b * v_b) / ((1/3) * M_bar * L_bar).

Step 6: Substituting the given values gives us: w_bar = - (0.009 kg * 0.21 m/s) / ((1/3) * 0.049 kg * 1 m) = -0.126 rad/s. The negative sign indicates that the bar rotates in the opposite direction to the bug's jump.

Step 7: The total kinetic energy of the system just after the bug leaps is the sum of the kinetic energy of the bug and the bar. The kinetic energy of the bug is (1/2) * m_b * v_b^2 and the kinetic energy of the bar is (1/2) * I_bar * w_bar^2.

Step 8: Substituting the given values gives us: KE_total = (1/2) * 0.009 kg * (0.21 m/s)^2 + (1/2) * ((1/3) * 0.049 kg * 1 m^2) * (-0.126 rad/s)^2 = 0.000198 J + 0.000328 J = 0.000526 J.

Step 9: This energy comes from the work done by the bug when it jumps off the bar.

Problem 2:

Step 1: We are given the equation for the angular position of the flywheel: 𝜃 = (2.0 rad/s^3) * t^3.

Step 2: To find the angular position at t1 = 2.0 s, we substitute t = 2.0 s into the equation: 𝜃 = (2.0 rad/s^3) * (2.0 s)^3 = 16 rad.

Step 3: To convert this to degrees, we use the conversion factor 180 degrees = π rad: 16 rad * (180/π) = 916.3 degrees.

Step 4: To find the angular position at t2 = 5.0 s, we substitute t = 5.0 s into the equation: 𝜃 = (2.0 rad/s^3) * (5.0 s)^3 = 250 rad.

Step 5: To convert this to degrees, we use the conversion factor 180 degrees = π rad: 250 rad * (180/π) = 14324 degrees.