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A typical small rescue helicopter, like the one in the Figure below, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?

The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)

Strategy

Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.

Solution for (a)

We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find E_r . The angular velocity ω for 1 r.p.m is

and for 300 r.p.m

The moment of inertia of one blade will be that of a thin rod rotated about its end.

The total I is four times this moment of inertia, because there are four blades. Thus,

and so The rotational kinetic energy is

Solution for (b)

Translational kinetic energy is defined as

To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is

Solution for (c)

At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:

Discussion

The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T , for the four-rocket propulsion system shown in the Figure below. The sled’s initial acceleration is 49 m/s ², the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.

A sled experiences a rocket thrust that accelerates it to the right.Each rocket creates an identical thrust T . As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight,w.The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger than scale.
Assumptions: The mass of the Sled remains steady throughout the operation

Strategy

Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.

Solution

Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with

Fnet is the net force along the horizontal direction, m is the rocket’s mass and a the acceleration.

We can see from the Figure at the top, that the engine thrusts add, while friction opposes the thrust.

Tt is the total thrust from the 4 rockets, Fnet the net force along the horizontal direction and Ff the force of friction.

Finally, since there are 4 rockets, we calculate the thrust that each one provides:

T is the individual Thrust of each engine, b is the number of rocket engines

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

In the wheelbarrow of the following figure the load has a perpendicular lever arm of 7.50 cm, while the hands have a perpendicular lever arm of 1.02 m.(a) What upward force must you exert to support the wheelbarrow and its load if their combined mass is 45.0 kg? (b) What force does the wheelbarrow exert on the ground?

(a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel’s axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone. (b) In the case of the shovel, the input force is between the pivot and the load, but the input lever arm is shorter than the output lever arm. The pivot is at the handle held by the right hand. Here, the output force (supporting the shovel’s load) is less than the input force (from the hand nearest the load), because the input is exerted closer to the pivot than is the output.

Strategy

Here, we use the concept of mechanical advantage.

Discussion
An even longer handle would reduce the force needed to lift the load. The MA here is:

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational.

(a) Calculate the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child).

(b) Calculate the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29e+11 m away. How does the force of Jupiter on the baby compare to the force of the father on the baby?

Father’s gravitational force on the baby is:

Jupiter’s gravitational force on the baby is:

(c) What should be the father’s weight, so that he exerts the same force on the baby as that of Jupiter? **
**this section is not included in the Reference material

Discussion

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

Dedicated to little Konstantinos