A. Questions on energy and power.
1. If a rower is rowing with a power output of work of 300 watts, how long does he have to row to burn off a big Mac. [700 Kcalories]? [remember that only 25% of the food energy is converted to work.] However, excercise for weight control works because when you are trained as an athlete, your body metabolism rate is raised even when you are idle.
A rower with a power output of 300 watts is burning stored energy at a rate which is 4X bigger [25% efficiency at delivering work] so is burning energy with power = 1200 watts. The big mac of 700 Kcalories = 700Kc*4186J/Kc = 2.9M Joules.
P =DE/Dt. Dt= DE/P = 2.9M/1200 = 2442 seconds = 41 minutes.
2. How much food must a climber eat to climb mount Rainer from sea level considering only the gravitational energy expended. ? Is this reasonable? h = 14,000' = 4200m
The gravitational energy = mgh = 70 * 10*4200 = 2.9MJoules. This is just what a big mac provides!
From the reports of climbers, much more input energy is required. The above is not a reasonable estimate of the energy requirements. The climbers spend more than one day usually with a sleep then a 0300 hrs ascent. There is the basal metabolism to feed for this time as well as the extra energy to supply heat for the body and probably the muscle tension from the nervousness of most people.
3. Use the energy argument to estimate that the maximum height of jumping is independent of the size of the athlete.
The gravitational energy required to reach a height, h . E= mgh
h = E/mg
Since the energy provided by the muscle is E = F*length. Assume that the body scales geometrically so that the length and width and thickness are proportional to a characteristic length , l. Also note that the force produced depends on the cross sectional area of the muscle [l^2] and the length is proportional to l. This means that the energy is proportional to l^3. Also since mass is proportional to volume or l^3 as well. We have that h is proportional to l^3/(l^3*g) or 1/g. The height of a jump does not depend on size. This is true for humans but also approximately true for species of different sizes. [dogs to horses jump about 1 m.]
4. Estimate how much energy is stored in the soles of some thick soled athletic shoes and compare it with the energy stored in the arch and achilles tendon. Does this explain why sprinters and jumpers use the thin soled track shoes and don’t use these thick soled shoes?
The soles of athletic shoes usually can compress by only a small amount under the weight of an athlete [1/8"??]. The energy used in compression is proportional to the compression length. E = F*d = 700* .003 = 2 Joules. This is considerably less than the energy stored in the arch and achilles tendon. Also the sole is not so elastic. [Try bouncing a shoe on it's sole] So it's better not to have an athletic shoe soaking up energy inelastically and also the weight of the sole will slow down the movements of the athlete.
5. How many pounds of muscle are required to launch a 70kg athlete on a 1 meter vertical jump?
For a 70kg To jump one meter requires E=mgh = 700J. Since the take off velocity is 4.5m/s, and the thrust is over about .2 m. The time of the thrust is
t =d/vavg =.2/2.25= 0.09sec. The power = 700/.09 = 7800watts = 10 HP. Since muscles produce power of rate=1/4 HP/pound, we need to have
weight = Power/rate= 10/(1/4)=40lb muscle!! A well developed athlete will have half of his or her weight in muscle. The 70kg athlete weighs 154 lb. This means that 1/2 the muscles have to be involved in the jump to one meter! Unfortunately, about half the muscles would be antagonists so the muscle mass would be twice that of the agonists or about 80 lb or ALL of the muscle mass in the athlete. The above calculation assumes that all of the jump comes from the thrust from the legs. We have also seen that the athlete does, indeed spread the power over a large part of the body and that the time for the jump is distributed over a bigger time with a reduced power. In any case, it's an exceptional athlete who would be able to jump that high.... more than training... genetics??
6. We have determined that the scaling law for lifts varies as m^(2/3) for the lift potential of athletes of mass,m. Lift ~ m^(2/3). Lift'/Lift = (m'/m)^2/3=(100/200)^(2/3) = 2^(2/3) = 1.6.
So Lift' = 1.6*Lift = 240 kg.
1. For an incoming ball with speed 90 mph, what must the racket speed be in order to deliver a 90 mph return? Can you feed of your opponents speed?
vball = 3/2vracket+ 1/2vball-in.
vracket =2/3[ vball-1/2vball-in] = 2/3[90-45]=30mi/hr.
This is quite a modest speed for the racket. Indeed, to generate your own service of 90mph would require a 60mph racket speed. Yes, one can feed off the speed of the incoming ball.
2. if a tackler and ball carrier approach head on each with a speed of 10m/s, what is the velocity of the two when the tackler holds on the the carrier? why?
If the two players have equal mass, they would have the same magnitude of momentum and in opposite directions. The net momentum would be zero before the collision and must be zero AFTER the collision. In this case, the two will have zero velocity after the tackle if they stick together in the tackling process.
3. For a massive completely elastic tennis racket, what is the speed of a struck ball if the incoming ball has speed 30 mph and the racket head is 30mph.
From the point of view of the massive racket head, the ball is coming at it with a speed of 60mph. With an elastic collision, the ball comes off the racket with a speed of 60mph as viewed by an insect on the racket.
From the player on the court, the ball comes off at 90 mph. v = 30+60
4. The runner has a hang time of 0.26 seconds while both feet are in the air. The foot comes down and bounces upward elastically, what is the velocity upwards as the foot leaves the ground? How high will the runner bounce up to?
If the hang time is 0.26 seconds, the downward flight takes 0.13 sec and the velocity before the collision is v = gt = 10*0.13 = 1.3 m/sec. The elastic collsion provides an upward velocity of 1.3m/sec upwards. The maximum height is calculated from
mgh= 1/2mv^2 or h = 1/2 v^2/g = 1/2 * 1.3^2/10 = 0.08m
5. If a baseball game were to be played in smaller fields, the home run potential may be so large as to make the game too much tilted to the offense. In Japan where games do take place in smaller fields, the game is played with a deader ball. Calculate the CoR for a ball for which the minimum homerun distance is changed from 90m to 75m.
The launch speed for a given bat speed must be reduced from 40m/s to 35m/s, a 12% decrease. Since the relationship of launch speed to CoR is given as:
Vlaunch = Mbat/(Mbat+Mball)*((Mball/Mbat - CoR)*(-Vball)+(1+CoR)*Vbat)
which is linear in CoR, the approximate change in CoR must be to reduce the Cor by 12% or from 0.55 to 0.48 for a higher speed collision. In Japan, I've seen some neighbourhood games played in a smaller field where the bat-ball collision sounds like a thud rather than a sharp crack.