The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [?Hint: ?Integrating Eq. (10.29) yields is called the angular impulse.]

Solution 92P Step 1 of 5: Given solid wood door has width 1m and height 2m and mass M=35 kg is hit by basketball at the center of the door with a force of 1500 N to the door for 8 ms as shown in the figure below. We need to calculate the angular speed of door after impact. Step 2 of 5: Given data, Mass, M=35 kg Distance from axis, x=0.5 m Applied force, F= 1200 N Time interval, t =0 and t = 8ms= 8× 10 s 3 1 2 We know that,for linear motion; impulse is equal to change in linear momentum. Similarly in case of rotational impulse it is equal to change in angular momentum. That is, 2 L = dt t[ ] 1 Since here the torque is due to the force applied by basketball That is = F x Step 3 of 5: Above equation becomes, t2 L = [F x dt] 1 3 Substituting t =1 and t = 82 10 s 8×103 L = [F x dt] 0 On integrating 8×10 s L = F|x t 0 Substituting F=1200 N , x =0.5 m 3 L = 1200 N (0.5 m)(8 × 10 s) L = 6 kg.m/s 2