1. Consider a ball on a fixed-length string being whirled in a vertical circular
ID: 1282549 • Letter: 1
Question
1. Consider a ball on a fixed-length string being whirled in a vertical circular path as shown in the diagram below.
(a) When the ball is at the top of the circle, what is the tension in the string? (Use the following as necessary: the length L of the string, the mass m, speed v of the ball, and gravitational acceleration g.) T =
(b) What minimum speed should the ball have at the top of the circle to continue in its circular motion? (Use any variable or symbol stated above as necessary.) vmin=
2. As you go above the Earth's surface, the acceleration due to gravity will decrease. Find the height above the Earth's surface where this value will be 1/110 g. m=
3. A geosynchronous Earth satellite is one that remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth's rotation). Such orbits are useful for communication and weather observation. Calculate the radius of such an orbit. Mass of Earth = ME = 5.97 multiply.gif 1024 kg. m =
4. Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 41.4 km/s, and v2 = 58.6 km/s respectively. (a) If the period of the first planet P1 is 770 years what is the mass of the star it orbits around? = kg (b) Determine the orbital period of P2. = yr
Explanation / Answer
using the angle theta measured with respect to the horizontal , going counterclockwise (as we normally do in trig) we have
Fc (centripetal force) = Tension + component of gravity in direction of tension
component of gravity in direction of tension = mg * sin theta , using a triangle argument is mg sin theta, therefore
Fc = T + mg sin theta
Also keep in mind that centripetal acceleration for constant velocity in a circle is v^2/L
so Fc = mass * acceleration = m*v^2/L
Therefore , an equation that will come useful later:
mv^2/L = T + mg sin theta
a) when ball is at bottom of circle tension is pointing upwards , in the same direction as centripetal force
b) weight always points down and the weight vector is equal to -mg so the answer is 5
c) centripetal force is always center seeking. And it is pointing up here towards the center so 5
d) the centripetal force Fc = T + mg sin theta, here theta = 3pi/2
Fc = T + mg sin(3pi/2)
Fc = T - mg
e) Fc = mv^2/L
f) mv^2 = T -mg
T = mv^2 + mg
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