A hammer of mass 8.00kg is attached with one end of a thin, light rod. A thrower

Question

A hammer of mass 8.00kg is attached with one end of a thin, light rod. A thrower accelerates the hammer from rest within four full turns(revolutions) and releases it at a speed of 21.0m/s. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20m, a) calculate the angular acceleration. b) the tangential acceleration c) the centripetal acceleration just before release. d) the net force being exerted on the hammer by the athlete just before release. e) the angle of this force with respect to the radius of the circular path

Explanation / Answer

mass of hammer m = 8 kg,speed of hammer linear v = 21.0 m/s, radius r = 1.2 m

from relation of linear and angular velocities V = rW ==> w = v/r = 21/1.2=17.5 rad/s

   using equations of motion like s = ut = 1/2 at^2

   angular displacement after completing 4 full turns is

   theta = 2pi*4 = 8pi =25.12 rad

   25.13 = 1/2 alpha t^2, and alpha is angular acceleration = theta /t

   25.13 = 0.5(25.13/t)t^2===> t = 2 s

now
a) angular acceleration alpha = 25.13/2 = 12.565 rad/s2

b) tangential acceleration is a = r *alpha = 1.2*12.565 = 15.078 m/s2

c) centripetal acceleration just before release is q = v^2/r = 21^2/1.2 = 367.5 m/s2

d) net force being exerted on the hammer by the athlete just before release.

   F= ma = 8*367.5 = 2940 N
e) gravity acting on it is Fg = mg = 8*9.8 = 78.4 N

net force is F = sqrt(78.4^2+2940^2) = 2941.0 N

angle theta = arc tan (78.4/2940)= 1.52 degrees

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