1) A thin rod (length = 1.79 m) is oriented vertically, with its bottom end atta

Question

1) A thin rod (length = 1.79 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

3) A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration 0.175 rad/s2. After making 2870 revolutions, its angular speed is 124 rad/s. (a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?

Explanation / Answer

1)

a) mgh = 1/2 mv^2
gh = v^2/2
2gh = v^2
v = (2gh)^.5
v = (2(9.81)(1.79))^.5
v = 5.926 m/s /1.79 = 3.31 rad/s

b) at horizontal the mass m will be accelerating at g or 9.81 m/s/s.
dividing by radius of 1.33 m gives
9.81 / 1.79 =5.48 rad/s/s

2)

a) You need to first convert 3.53 rev/s to rad/s to use the equation given. Do this by using the equation below:

rad/s=2*pi*(rev/s)

2*pi*3.53=22.18 rad/s =w

Now use the equation given in the problem:

vT=rw

vT = .0568*22.18 = 1.26 m/s

b) Use the tangential speed found in part (a) to solve part (b).

1.26 =0.0439*w ==> 1.26/0.0439=w

w = 28.70 rad/s = 4.57 rev/s

3)

a) ² = ² + 2 ( =124 rad/s, =2870*2pi, = 0.175 rad/s2 )
= (² - 2)
= 95.21 rad/s

(b) = + t
t = ( - )/
t = 164.51s

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