Circular Motion Review

From TuHSPhysicsWiki
Jump to: navigation, search

Main Page > Solutions to Worksheets > Circular Motion Review

1. Bob twirls a 0.615 kg mass at a constant speed in a 34.0 cm radius circle at a speed of 1.95 m/s. What is the centripetal acceleration? What force in what direction is he exerting at the top and the bottom

We're looking to find the centripetal acceleration in this problem. We know the radius (34.0 cm), the velocity (1.95 m/s), and the mass (0.615 kg). So the formula for acceleration is a = \frac {v^2}{r}

so a = \frac {(1.95\ m/s)^2} {.340\ m} = 11.18382\ m/s/s
So now we know the acceleration, we must find the force at top and bottom of the circle.

To solve this, use newton's second law (F = ma). F is the sum of the forces acting on the mass, and a is the centripetal acceleration in this case. Remember that at the top of the circle, you are accelerating down (toward the center), and at the bottom you are accelerating up (again toward the center)

First - let's find the force of gravity on the 0.615 kg mass:
F = ma = (0.615 kg)(9.81 N/kg) = 6.03315 N


So at the bottom we have the unknown force F, and weight effecting an upward (+) acceleration of 11.1838 m/s/s:

\ F = ma
\ F - 6.03315 = (0.615\ kg)(+11.1838\ m/s/s)
\ F = +12.9112\ N = 12.9 N\ up


At the top we have the unknown force F and the weight effecting a downward (-) acceleration of 11.1838 m/s/s:

\ F = ma
\ F - 6.03315 = (0.615\ kg)(-11.1838\ m/s/s)
\ F = -0.84490\ N = 0.845\ N\ down


Notice that the two F = ma expressions are different only in that the acceleration changes direction, everything else is identical

Table of Contents


2. Marlene twirls a 2.50 kg mass in a 62.0 cm radius vertical circle at a constant speed. At the bottom of the circle she needs to exert an upward force of 56.4 N of force. What is the centripetal acceleration? What force in what direction is exerted at the top?

For this one we will need to begin with the F = ma stuff.

So at the bottom, we have a force of 56.4 N up (+) and the weight of the mass (2.50 kg)(9.81 N/kg) = 24.525 N down (-) causing an upward (centripetal) acceleration:

\ F = ma
\ +56.4\ N - 24.525\ N = (2.50\ kg)a
\ a = +12.75\ m/s/s

So the centripetal acceleration is 12.75 m/s/s


So now at the top we have an unknown force F and the weight (24.525 N down) effecting a downward (-) acceleration of 12.75 m/s/s:

\ F = ma
\ F - 24.525 N = (2.50\ kg)(-12.75\ m/s/s)
\ F = -7.35\ N

So the downward force is about 7.35 N.

Table of Contents


3. Red Elk twirls a 15.0 kg mass in a 1.50 m radius vertical circle. He is so clever that he can do this at a constant speed. At the top, he is pulling down on the mass with a force of 81.0 N. What is the centripetal acceleration of the mass? What is the force he must exert at the bottom?

For this one we will need to begin with the F = ma stuff.

So at the top, we have a force of 81.0 N down (-) and the weight of the mass (15.0 kg)(9.81 N/kg) = 147.15 N down (-) causing a downward (centripetal) acceleration:

\ F = ma
\ -81.0\ N - 147.15\ N = (15.0\ kg)a
\ a = -15.21\ m/s/s

So the centripetal acceleration is about 15.2 m/s/s


So now at the bottom we have an unknown force F and the weight (147.15 N down) effecting an upward (+) acceleration of 15.21 m/s/s:

\ F = ma
\ F - 147.15 N = (15.0\ kg)(+15.21\ m/s/s)
\ F = 375.3\ N

So the upward force is about 375 N.

--Cmurray 11:50, 11 February 2009 (PST)

Table of Contents


4. At what distance from the center of the Earth is the force of gravity 13.5 N on an 18.0 kg mass?

So this is a basic plug and chug. But be careful how you both plug and chug:

F = \frac{Gm_1m_2} {r^2}

Plugging in numbers:

13.5\ N = \frac{(6.67\times10^{-11}\ Nm^2kg^{-2})(5.97\times10^{24}\ kg)(18.0\ kg)} {r^2}

And if you can do math (don't forget to square root) you will get that r = 23041961.72 which is 2.30 x 107 m

--Cmurray 11:57, 11 February 2009 (PST)

Table of Contents


5. What is the orbital velocity 2.00x106 m from the center of the moon? What is the period of your motion here?

So in orbit, you just set the centripetal force equal to gravity. Use these formulas:

\ F = ma

a = \frac{v^2}{r} = \frac{4\pi^2r}{T^2}

F = \frac{Gm_1m_2} {r^2}


In this case let's use the acceleration with velocity in it, as that is what we are solving for

\frac{Gm_1m_2} {r^2} = \frac{m_2v^2}{r}

Where m2 is the satellite mass. (it is the one accelerating)


Now if you solve that for v you get:

v = \sqrt{\frac{Gm_1}{r}} = \sqrt{\frac{(6.67\times10^{-11}\ Nm^2kg^{-2})(7.36\times10^{22}\ kg)}{(2.00\times10^6\ m)}} = 1566.70\ s

So the orbital velocity is about 1570 m/s


Now to find the period we could start over and solve

\frac{Gm_1m_2} {r^2} = \frac{m_24\pi^2r}{T^2}

for T which really looks like not so much fun. A simpler approach would be to just remember (it is not in the data packet) the relationship between circumference velocity and period. Since velocity is distance divided by time, and the distance is 2 pi r, and the time is the period, you get:

v = \frac{2\pi r}{T}

so:

T = \frac{2\pi r}{v} = \frac{2\pi (2.00\times10^6\ m)}{(1566.70\ m/s)} = 8020.8988...s

So the period is about 8020 s

--Cmurray 12:11, 11 February 2009 (PST)

Table of Contents


6. What is the period of motion at a distance of 1.496x1011 m from the center of the sun? Give it in seconds and in days. (3600 seconds to an hour, 24 hours per day…)

Since this involves the period, you would choose the orbital condition:

\frac{Gm_1m_2} {r^2} = \frac{m_24\pi^2r}{T^2}

Solving for T gives:

T = \sqrt{\frac{4\pi^2 r^3}{Gm_1}} = \sqrt{\frac{4\pi^2 (1.496\times10^{11}\ m)^3}{(6.67\times10^{-11}\ Nm^2kg^{-2})(1.99\times10^{30}\ kg)}} = 31556425.51 s

Which is about 3.156x107 s. If you divide this by 3600 to get hours, and 24 to get days the result is 365.2364064 days

So it is about 365.24 days

--Cmurray 12:44, 11 February 2009 (PST)

Table of Contents


7. You are orbiting a planet with an orbital radius of 7.34x106 m at a velocity of 7350 m/s. What is the mass of the planet?

We need to find the mass of the planet. We know that the planet has a radius of 7.34x106 m, and that we have a velocity of 7350 m/s.
You would use this condition of orbit:

\frac{Gm_cm_s} {r^2}=\frac{m_s v^2}{r}

Solving for m:

m_c = \frac {v^2 r}{G} = \frac {(7350\ m/s)^2 (7.34\times10^6\ m)}{(6.67\times10^{-11}\ Nm^2kg^{-2})} = 5.9449\times10^{24}\ kg

So the mass is 5.94x1024 kg

\infty\tilde{a}\ddot{a}\hbar \;\ell \;\infty --10lucasa 09:29, 13 February 2009 (PST)


Table of Contents


8. What is the centripetal acceleration at the edge of a record 0.19 m from the center of revolution when the record is turning at 78 RPM?

To find the period simply divide the 60 seconds in a minute by the RPM:

T = \frac{60 \frac{sec}{min}}{78 \frac{rev}{min}} = .76923... \frac{sec}{rev}

and now just plug this into the formula:

a = \frac{4\pi^2r}{T^2} = \frac{4\pi^2(0.19 m)}{(.76923... s)^2} = 12.676... m/s/s

a = 13 m/s/s

Table of Contents