When a turntable rotating at 𝜔𝑡 = 34.4 𝑟𝑒𝑣𝑚𝑖𝑛 is shut off, it comes to rest in 𝑡 = 25.0 𝑠. Assumingconstant angular acceleration, find the angular acceleration, the average angular velocity of theturntable, and the number of revolutions it makes before stopping. (3 sig. figs)Problem 4A thin, uniform metal bar, 𝐿𝑏𝑎𝑟 = 2.00 𝑚 long and weighing 𝐹𝑏𝑎𝑟 = 90.0 𝑁 , is hangingvertically from the ceiling by a frictionless pivot. Suddenly it is struck 𝐿ℎ𝑖𝑡 = 1.50𝑚 below theceiling by a small 𝑚𝑏 = 3.00 𝑘𝑔 ball, initially horizontal at 𝑣𝑖 = 10.0 𝑚𝑠 . The ball rebounds inthe opposite direction with a speed of 𝑣𝑓 = 6.00 𝑚𝑠 . Find the angular speed of the bar just afterthe collision. During the collision, why is the angular momentum conserved but not the linearmomentum? (3 sig figs)

Let's break down the problem into two parts and solve them separately.

Part 1: The turntable is rotating at 34.4 rev/min which is equal to 34.4*2π rad/min = 216.2 rad/min. To convert this to rad/sec, we divide by 60 to get 3.60 rad/sec.

The turntable comes to rest in 25.0 sec. Since the final angular velocity (ωf) is 0 rad/sec, we can use the equation for angular acceleration (α) which is (ωf - ωi) / t. Substituting the given values, we get α = (0 - 3.60 rad/sec) / 25.0 sec = -0.144 rad/sec².

The average angular velocity (ωavg) is (ωi + ωf) / 2 = (3.60 rad/sec + 0) / 2 = 1.80 rad/sec.

The number of revolutions it makes before stopping can be found using the equation θ = ωit + 0.5α*t². Substituting the given values, we get θ = 3.60 rad/sec * 25.0 sec + 0.5 * -0.144 rad/sec² * (25.0 sec)² = 90.0 rad. Since 1 revolution is 2π rad, the number of revolutions is 90.0 rad / 2π = 14.3 revolutions.

Part 2: The angular speed of the bar just after the collision can be found using the principle of conservation of angular momentum. The initial angular momentum of the system is the momentum of the ball which is mvLhit = 3.00 kg * 10.0 m/s * 1.50 m = 45.0 kgm²/s. The final angular momentum of the system is the momentum of the ball and the bar which is mvLhit + Iωf. The moment of inertia (I) of the bar is mL²/3 = 90.0 N * (2.00 m)² / 3 = 120.0 kgm². Setting the initial and final angular momentum equal to each other and solving for ωf, we get ωf = (45.0 kgm²/s - 3.00 kg * 6.00 m/s * 1.50 m) / 120.0 kgm² = 0.300 rad/sec.

During the collision, the angular momentum is conserved because there are no external torques acting on the system. However, the linear momentum is not conserved because the force exerted by the bar on the ball is not equal and opposite to the force exerted by the ball on the bar.