Chapter 8 Rotational Equilibrium and Rotational...
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Chapter 8 Rotational Equilibrium and Rotational Dynamics HW# 1, 3, 7, 9, 11, 17, 19, 31, 33, 37, 39, 43, 51, 55, 61
Transcript of Chapter 8 Rotational Equilibrium and Rotational...
Chapter 8Rotational Equilibrium
and Rotational DynamicsHW# 1, 3, 7, 9, 11, 17, 19, 31, 33, 37, 39, 43, 51, 55, 61
See figures on page 241
r is the moment arm (lever arm)
Example problem 8.1
What to do when r and F are NOT perpendicular.τ = r F sin θθ is the angle between the r and F vectors.
See figure 8.4, 8.5
The vector component of the force that is perpendicular to the moment arm (r) provides the torque.
The vector component of the force that is parallel to the moment arm produces no torque.
Center of Gravity
See figure 8.10 for a method in experimentally determining the center of gravity.
Important detail: The center of gravity of an object does not have to be where the object has any mass.
Weighted forearm example. Page 250
We want to find the upward force exerted by the biceps and the downward force exerted by the humerus.
Forearm is in equilibrium.Apply condition of the net force on forearm equals zero.
Apply condition of the net torque on the forearm equal zero.
The angular acceleration of a rigid body is proportional to the net torque acting on it.
The proportionality factor is the moment of Inertia.
Στ = I α
On page 257, we can see what the moment of inertia is for a few commonly shaped objects.
Examples 8.10 and 8.11
Example problem 8.85
Iron Cross: Gymnast weighing 750 N.Each ring supports half the weight.Fs is force exerted by the shoulder joint.Fm is the force exerted by the pectoralis major and latissimus dorsi.Fm acts 4 cm from the shoulder at a downward 45 degree angle.The arm is 70 cm long.Estimate force Fm.
Apply the condition of Στ = 0,Where the pivot point is the shoulder.
w = weight = 750 N w/2 = 375 N
(w/2) (70 cm) – (Fm sin 45) (4 cm) = 0
Fm = 12.4w = 9300 N
8.52 Bucket on a spool.
Use conservation of energy to determine the angular speed of the spool, after the 3 kg bucket has fallen 4 meters, starting from rest.
mb = 3 kgms = 5 kgrs = 0.6 m
Conservation of EnergyKEi + PEi = KEf + PEf
Pick the zero of potential energy line to be where the bucket is lowered to.Initial kinetic energy = 0
Final kinetic energy is from the motion of the bucket and the spinning of the wheel.
0 + mbgh = ½ mbvb2 + ½ Isω2
