Section 1
1.
What two factors (hint: they are anthropometric characteristics) affect the magnitude of terminal velocity for a falling object such as a cat?
The two factors that affect terminal velocity for a falling object is the surface (body) area and the mass of the object
2.
Explain how a smaller mass-to-area ratio results in a lower terminal velocity for a falling cat, relative to a fallling human?
Since a cat cannot change its mass during a fall, it changes its surface area so that it can increase its air resistance and decrease its terminal velocity. According to the article, once a cat hits terminal velocity, it relaxes its muscles and spreads its limbs horizontally like a flying squirrel, thus changing the surface area. Humans could try and imitate the cat by spreading out their limbs. Doing so would give them a lower terminal velocity.
However, the greater mass to area ratio in humans, gives the human a greater terminal velocity when compared to a cat.
Section 2
1.
During the Olympic event of hammer toss, an athlete gets a 7.25 kg hammer swinging at
570º/s. If the hammer is at the end of a 1.22 m wire what is the minimum amount of force needed to hold on to the hammer without it slipping out of the athletes hands.
First convert the degrees to radians. 53.7º = 1 radian, so 570º = 9.95 radians.
Next, use the centripetal force equation F c
= mrω
2 to find the force needed to hold onto the hammer. F c
= 7.25 kg * 1.22 M * (9.95rad)
2
. F c
= 875.26 N
2.
A figure skater jumps into the air with an I = 8.3 kg·m^2 and angular velocity of 7 rad/s. What will be her new angular velocity when she tucks in her arms and changes her moment of inertia to 4.3 kg·m^2?
First, calculate her initial angular momentum using the equation H = Iω. H = 8.3 * 7. H =
58.1. Because momentum is conserved during flight, use the calculated momentum and the new moment of inertia to calculate the new angular velocity, using the same equation as above. 58.1 = 4.3 * ω. Solving for ω, divide both sides by 4.3 and
ω = 13.5 rad/sec.

3.
The mass of a baseball bat is typically 1 kg and the radius of gyration is 0.62 m. A certain player normally swings a regular bat with an angular velocity of 420º/s, but interestingly during one game, he is suddenly able to swing with a bat speed of 530º/s. It is discovered that the bat he used was corked (hollowed out) in order to increase bat speed. Assuming the bat was the regular length and he swung with the same momentum, what was the mass of the corked bat?
Everyone gets full credit for this question.
4.
The bicep muscle inserts on the radius (bone) 2 cm from the elbow at a 35º angle. What force must the bicep exert to create an angular acceleration of 100º/s^2 in the arm if the arm has a moment of inertia equal to .14 kg·m^2?
First, draw a picture to indicate the known variables.
Fb
35 º
.02m
Next, convert the degrees to radians for angular acceleration. Remember, 53.7º = 1 radian, so 100º = 1.75 radians. Knowing that ∑T = Iα, and that T = F* ⊥ , we know that F
* ⊥ = Iα. In order to solve for the force, we need to solve for the perpendicular distance of that force. This is done by using sin of the angle and the known distance
(the hypotenuse). So, ⊥ = sin(35) * .02. ⊥ = 0.0115m. Now plug the known values in to solve for the force of the biceps. F = (0.14 * 1.75)/ 0.0115. F b
= 21.3 N