Question

Problems 3: The 5-kg disk is pinned with a frictionless pin at O A 1 bullet is fired with v-800 m/s into the edge ot the disk, and the embedded in its edge. Ignore the dimension of the bullet. The gravity acts downward. bullet becomes embedded in it. Answer the following questions just after the bullet becomes 0.2 m 3.1 The angular velocity of the disk israds (A) 2.01 (B) 3.99 (C) 6.32 (D) 4.65 (E) 5.32(F) 7.39 nc 3.2 The velocity of mass center G of the disk is m/s m's (A) 0.798 (B) 1.82 (C) 0.931 (D) 1.06 E) 3.02 (F) 4.31 3.3 If the bullet becomes embedded in 0.005 s, the magnitude of impulsive force is kN. (A) 1.20 (B) 1.80 (C) 2.01 (D) 4.32 (E) 1.40 (F) 1.60 3.4 The velocity of bullet is m/s. (A) 1.50 (B 1.85 (C) 1.98 (D) 1.32 (E) 1.13 (F) 2.31

Answer 1

3.1 Calculate the velocity of combined mass system just after bullet embedded in the disk. (Using law of conservation of linear momentum) mv V- m+M (0.01)(800) (0.01+5) 1.5968 m/s Calculate the position of center of mass of combined mass system from center of disk. M-0+m(0.2) M +m 0.2m M+m 0.0003992 m Calculate the angular velocity of the disk.

1.5968 (0.0003992) +(0.2) -7.984 rads 7.98 rad/s The calculated value is closer to the value in option (F). Therefore, option F is correct 3.2 The velocity of the mass center G of the disk is, V-1.5968 m/s 1.60 m/s The calculated value is closer to the value in option (B). Therefore, option (B) is correct 3.3 Calculate the impulsive force.

mu-T (0.01) (800-1.5968) (0.005) -1596.81N 1.60 kN Therefore, option E is cormrect. 3.4 Bullet gets embedded in the disk. Therefore, bullet velocity will remain same as the velocity of the combined mass system after the impact. Therefore, the velocity of the bullet is, V = 1.5968 m/s 1.60m/s The calculated value is closer to the value in option (A). Therefore, option (A) is correct