Question

Biomechanics!

A) A picture holding a ball, rotates his arm at an angular velocity of 50 radians/sec. If his arm is 0.8 meters in length calculate how high a ball would reach if the ball was released vertically. (HINT: first calculate the initial velocity then treat it like a vertical projection problem)?

B) Skater on team A, (mass 80 kg), is moving at a velocity of 12 m/sec collides with skater on Team B, (mass 75 kg) moving in the opposite direction. The two now move together at a velocity of 3 m/sec, in the direction that player on Team B was originally moving. What was the team B skater original velocity?

Answer 1

A. v= rw

Where v = linear speed, r= radius, w=angular velocity

v=0.8*50= 40 m/s

This speed will become the initial velocity of projection when projected upwards

use v^2= u^2 +2as , s=displacement,u=starting velocity ,v=final velocity,a=acceleration(In this case its acceleration due to gravity -9.8 m/s^2 , At the top most height it will stop for a moment so final velocity =0

0^2= 40^2 -2*9.8*s

s= 81.63 m

B. Lets apply law of conservation of linear momentum

m1u1 + m2u2 = m1v1 + m2v2 (m=being masses, u=starting velocities ,v=final velocities) ,We take velocity of A to be positive and velocity of B to be negative.

80*12 -75 u2 = -80*3-75*3

u2= 19 m/s