course Phy 121
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Fcent = v^2 / r
2 m/s^2 / 4 m = .5 m/s^2
If the object has mass 85.99 kg, how much force is required to provide the acceleration?
F = m * a
F= 85.99 kg * .5 m/s^2
F= 43 Newtons
The problem wasn't included but, assuming the problem asked for the force on an 85.99 kg object moving at sqrt(2) m/s on a circle of radius 4 m, the solution looks good.
However note that the units of v^2 are m^2/s^2, which when divided by m gives you m/s^2.
Problem Number 2
A massless disk is constrained to rotate about an axis through its center and perpendicular to its plane. Iron rods are shaped into circles whose radii are 7.9 meters, 15.8 meters and 23.7 meters. These circles are secured to the disk, concentric with it. The rods have mass density 10.05857 kilograms/meter.
Find the angular acceleration that will result from the application of a torque of 1044 meter Newtons.
Alpha = ?au / (mr^2)
= 1044 meter N / (10.05857 kg / m * 47.4 m)
= 2.08 N / kg/m
I am not sure if I did this problem correctly.
You have to find the mass of each circle, and compute m r^2 for each. Then add up the m r^2 to get moment of inertia I.
Then alpha = tau / I.
Problem Number 3
Using proportionality, the acceleration of gravity at the surface of the Earth and the fact that the radius of the Earth is approximately 6400 km, use proportionality to find:
The field strength at twice the radius of the Earth from its center.
A = gm / r^2
A = (6.6742*10^-11 N m^2 kg^-2) * 5.9742 ?1024 kilograms / 12800 km^2
= 2433655.129 kg/m^2
N m^2 kg^-2 * kilograms / km^2 does not give you kg / m^2. Among other things, km^2 must be changed to m^2 to be compatible with the m^2 in the numerator.
The strength at 4 times the radius of the Earth from its center.
A = (6.6742*10^-11 N m^2 kg^-2) * 5.9742 ?1024 kilograms / 25600 km^2
= 608413.78 kg/m^2
?The strength at a distance of 28100 kilometers from its center.
A = (6.6742*10^-11 N m^2 kg^-2) * 5.9742 ?1024 kilograms / 28100 km^2
= 504970.88 kg/m^2
?The distance from Earth's center where the gravitational field of the Earth is half its value at the surface.
A = (6.6742*10^-11 N m^2 kg^-2) * 5.9742 ?1024 kilograms / 3200 km^2
= 38938482.07 kg/m^2
N m^2 kg^-2 * kilograms / m^2 = (kg m/s^2) * m^2 * kg^-2 * kg / m^2 = (kg m/s^2) * m^2 * kg / ( kg^2 * m^2) = m/s^2.
The procedure is right but the units are not; it is especially important to change km to m.
Problem Number 4
A radian is the angle defined by a sector of a circle for which the arc of the circle is exactly as long as the radius of the circle.
?Since the circumference of a circle is 2 `pi r, it is possible to make 2 `pi sectors each with an arc of length r.
?So there are 2 `pi radians in a circle.
There are also 360 degrees in a circle.
?How many degrees therefore correspond to an angle of 1 radian?
57.3 degrees
?How many degrees therefore correspond to `pi , `pi /2 and `pi /6 radians?
?i = 180 degrees
?i / 2 = 90
?i / 2 = 30
Problem Number 5
Imagine that you are orbiting a neutron star whose mass is 43 * 10^30 kilograms at a distance of 20 kilometers from its center.
?Determine the average force gradient (force change per unit of distance), in Newtons per meter, for one kilogram masses between 20 and 20+1 kilometers from the center of the neutron star.
I know that F = m*a. I am not sure what to do.
?Use this gradient to estimate the difference you would therefore expect between one kilogram of your left shoulder and one kilogram of your right shoulder, assuming that one shoulder is .39 meter further from the star than the other.
Determine the average velocity gradient, in (m/s) per meter, between orbits at 20 km and at 20+1 km from the center.
?Determine how long it would therefore take one shoulder to move ahead of the other by one complete revolution.
See the solution to a problem of this type in the Introductory Problem Sets. I believe it is the last problem on gravitation. Let me know if you have questions.
Problem Number 6
An object of mass 680 kilograms is in circular orbit about a planet of mass 9 *10^ 26 kilograms at a radius of 20000 kilometers? How long will it take the object to orbit the planet?
This seems so simple but I can? figure it out.
Find the gravitational force using G M m / r^2.
Set this equal to the centripetal force m v^2 / r.
Solve for v.
The circumference is 2 pi r. Divide that distance by v to find the time required for an orbit.
This problem is also from the Introductory Problem Sets, and if you need more the details you can find them there. If that solution doesn't help, let me know what you do and do not understand and I'll be glad to clarify.
Problem Number 7
A mass of 7.599 kilograms is constrained by a massless rod to move in a circle of radius .8 meters. A torque of 9.199 meter Newtons is applied to the system, which is initially at rest.
?What force on the 7.599 kilogram mass, directed perpendicular to the rod, is equivalent to this torque?
tau = F*r
tau / r = F
9.199 m Newtons / .8 m = 11.5 Newtons
?What acceleration would this force give the mass?
F = m*a
F / m = a
= 11.5 Newtons / 7.599 kg
= 1.51 Newtons/kg
This is 1.51 (N m/s^2 ) / kg = 1.51 m/s^2.
?What angular acceleration would this correpond to?
Alpha = ?au / (mr^2)
Alpha = 9.199 m Newtons / (7.599 kg * .8 m^2)
= 1.89 Newton / kg / m
A Newton is a kg m/s^2.
Torque is not in Newtons but in meter * Newtons.
So you are dividing kg m/s^2 * m by kg m^2. This gives you
kg m^2 / s^2 / ( kg m^2) = 1 / s^2, or s^-2.
Find the quantity `tau / (mr ^ 2), where `tau is the torque, m the mass and r the radius of the circle.
?What is your result and what is its significance?
1.89 Newton / kg / m is the angular acceleration
See what you get with the above corrections.
Problem Number 8
How much paint is applied per square meter if 10 gallons of paint are uniformly spread out over the surface of a sphere of radius 2.1 meters?
Pi*r^2
Pi * 2.1^2
13.9 meters
Pi * (2.1 m)^2 = 13.9 m^2. The unit m is part of r and since r is squared, the unit has to be squared.
Also you should recognize that area is not measured in meters, but in m^2.
10gallons / 13.9 meters
= .72 gallons / meter
.72 gal/m^2<>
?If the paint is applied over a sphere of double the radius, by what factor does the amount per square meter change?
Pi*4.2meters^2
=55.4 meters
10 gallons / 55.4 meters
= .18 gallons/meter
.18 gal/m^2 would be correct.
By what factor has the amount per square meter changed?
?If the paint is applied over a sphere of four times the radius, by what factor does the amount per square meter change?
Pi*8.4 meters^2
=221.67 meters
10 gallons/221.67 meters
= .05 gallons/meter
.05 gal/m^2 would be correct, but does not have enough significant figures to serve as a basis for comparison.
By what factor has the amount per square meter changed?
If the paint is applied over a sphere of radius 17 meters, what is the factor by which the amount per square meter changes?
Pi*17 meters^2
=907.92 meters
10 gallons / 907.92 meters
= .011 gallons/meter
Again, what is the factor?
Do refer to the Intro Problem Sets for more detail on most of these problems.
You're not in bad shape; hopefully my notes will help. Let me know if you have questions.