A 230-kg beam 2.7 m in lenght slides bradside down the ice with a speed of 18 m/

Question

A 230-kg beam 2.7 m in lenght slides bradside down the ice with a speed of 18 m/s. A 65-kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion.

(a) How fast does the center of mass of the system after the collision?

(b) With what angular velocity does the system rotate about its CM?

A 230-kg beam 2.7 m in lenght slides bradside down the ice with a speed of 18 m/s. A 65-kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion. How fast does the center of mass of the system after the collision?

Explanation / Answer

a . Applying conservation of linear momentum we ve ..

Mv + 0 = (M + m)v

where v is the velocity after collision...

(230kg)(18m/s) = ( 230kg + 65kg)v

hence v = 14.03 m/sec

b . in the second question we need to find the location of centre of mass of the beam..

using the concept that angular momentum is conserved during collision

hence

d = {m/2l}(m+M)

= {65/2*2.7}(65+230) = 0.04 m

hence applying parallel axis theorem we ve

230*18*0.04 = 230*2.7^2/12 + 230*(0.04)^2 + 65*(1-0.04) w

hence w from here

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