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A) A grinding wheel starts from rest and attains a constant angular acceleration of

5.0 rad /s^−2. Calculate the acceleration at a point 1.0 m from the axis at t = 6.0 s.
b) An ice skater is spinning abo How can her
angular momentum be changed? How can her angular velocity be changed without
changing her angular momentum?



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R = 1.0 m

Angular acceleration = α  = 5.0 rad/sec²
Instantaneous Angular velocity = ω(t) = α t = 5 t  rad/sec

Instantaneous Velocity v(t)  of the point at 1 m from the center of grinding wheel
                 = r  ω(t)  = 1 * 5 t
      v (t) = 5 t  m/sec

Instantaneous centripetal acceleration = a_r = v² / r = 25 t² / 1 = 25 t²
         this is directed towards the center along the radius

Instantaneous linear (tangential) acceleration = r α
          =>  a_t  = 1 m * 5 rad/sec² = 5 m/sec²

Instantaneous resultant acceleration = a = √[ a_r² + a_t² ]
           a = √ [ 25 t² + 5² ]
     a  =  30.41 m/sec²  at 6.0 sec

the direction of the angular velocity : tan⁻¹ [ a_t / a_r ]
             = 1.59 deg. with the radius.

angular momentum = L = m v r  = I ω
         ω = m v r  /  I ,        where I is the moment of inertia.

Moment of inertia is increased by spreading the hands out...spreading the body out more.   So the radius of gyration increases. 

When the moment of inertia decreases, the angular velocity ω increases.  When the skater curls himself/herself inside and sits, then his/her radius of gyration decreases.  So moment of inertia decreases.  So the skater spins very fast.

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