**Centripetal stuff**

G ~= 6.6726 x 10^-11 m^3/kg*s^2

**1.** Uniform circular motion – when an object is moving
in a circle at a constant speed v. The magnitude of the velocity remains constant
in this case, but the direction of the velocity is continuously changing as
the object moves around the circle.

Centripetal acceleration – The acceleration that points toward the center
of the circle, also known as radial acceleration

** 2.** F(centripetal) = (mv^2)/r F(centripetal) = F(tension) –
F(normal)

**
3.** Centripetal force is the force that is always pointing inwards towards
the center of the circle. Centrifugal force is the imaginary feel that you’re
being thrown outwards.

5.

F = G(m1*m2)/r^2, r being the distance between m1 and m2.

**Problem 1**

For a spacecraft going from the earth toward the sun, at what distance from the earth will the gravitation forces due to the sun and the earth cancel? Earth mass: 5.98x10^24 kg, Sun's Mass: 1.99x10^30 kg, earth-sun distance: 150x10^6 km.

Start off be converting the earth-sun distance to meters.. => 1.50 x 10^11 m

Next, write the expression for net force on the spacecraft.

F_net = 0 = F_earth-space + F_sun-space.

F_earth-space = -F_sun-space.

Since we are only concerned with the magnitude of the force, we forget the -.

G(m_space*5.98x10^26kg)/r^2 = G(m_space*1.99x10^30 kg)/(1.50 x 10^11 m - r)^2

m_space cancels, G cancels.

5.98x10^26 kg / r^2 = 1.99x10^30 kg / (1.50x10^11 m - r)^2

you end up with a quadratic..

r ~= 2555943934.83m or -2646121787.27m

r can't be negative (the distance from spacecraft to sun will exceed the distance between earth and sun).

r ~= 2555943934.83m

**Problem 2**

A car moving at a speed of 26 m/s enters a curve that describes a quarter turn of radius 124 m. The driver gently applies the brakes, giving a constant tangential deceleration of magnitude 1.2 m/s^2.

a) Just before emerging from the turn, what is the magnitude of the car's acceleration?

b) At that same moment, what is the angle q between the velocity vector and the acceleration vector? Please enter your answer in degrees.

a) Lets just assume that the car is going north and reaches the quarter turn. The distance that it will be accelerating at -1.2m/s^2 can be found by finding the circumference of the would be circle divided by 4, 62pi. Now we use v^2=v0^2 +2ad to find the v^2, the final velocity after undergoing acceleration. This is 208.6192(m/s)^2 . We substitute this into the tangential acceleration equation, a = v^2/r and arrive at a = 1.682m/s^2. That is our tangential acceleration pointing towards the center of the turn. The other component of acceleration is pointing to the right. So, we resolve the vector => sqrt(1.2^2 +1.682^2) and get 2.066522m/s^2 as our acceleration.

b) The angle between the velocity vector and the acceleration vector can be found by the following. First determine the angle that the centripetal acceleration makes with the deceleration, arctan(1.682/1.2). Next, we subtract that from 90 to get the angle to the vertical. And finally, we add 90 to make the angle with the velocity vector: 125.505degrees

**Problem 3**

In an amusement park ride, a small child stands against the wall of a cylindrical room that is then made to rotate. The floor drops downwards and the child remains pinned to the wall. If the radius of the device is 2.15m and the relevant co. of friction is 0.40, what minimum speed to keep him pinned? (Thanks degoo_)

/\friction

||

||--> centripetal force, normal force on child is opposite in direction.

||

||

\/ gravity

Here we have the normal force to the wall of the room. It is equal to the centripetal force. Fnorm = mv^2/r.

The force of friction is then mu*mv^2/r. It will be equal to gravity if we want to find the minimum speed to keep him pinned.

mu*mv^2/r = mg, the mass term drops out of the equation like the floor of the room.

mu*v^2/r = g, v = sqrt(rg/mu).

Substitute the knowns in and arrive at v = 7.26m/s.