Free Online Flashcard Deck

5 - Circular Motion and Gravitation Free Online FlashCards

Study 5 - Circular Motion and Gravitation with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

Why does constant-speed circular motion involve acceleration?

Back

Yes. Its velocity changes direction continuously, so its acceleration points inward even though its speed remains constant.

02
Front

What is the direction of tangential velocity?

Back

The instantaneous velocity is tangent to the circular path and perpendicular to the radius from the center to the object.

03
Front

Is centripetal force a separate physical interaction?

Back

No. Centripetal force is the net inward radial force, which may be supplied by tension, friction, gravity, the normal force, or combinations of these.

04
Front

What is the period of circular motion?

Back

The period, TT, is the time required for one complete revolution.

05
Front

What equations relate angular speed, frequency, and period?

Back

Angular speed is related to frequency by ω=2πf\omega=2\pi f, or to period by ω=2πT\omega=\frac{2\pi}{T}.

06
Front

How are circular-path speed, radius, period, and frequency related?

Back

For uniform circular motion, the speed is v=2πrT=2πrfv=\frac{2\pi r}{T}=2\pi rf.

07
Front

What accelerations occur in nonuniform circular motion?

Back

It has both radial and tangential acceleration: a⃗=a⃗c+a⃗t\vec a=\vec a_c+\vec a_t. The tangential component changes speed, while the radial component changes direction.

08
Front

What is the maximum speed on a level curve?

Back

The maximum speed is vmax⁡=μsgrv_{\max}=\sqrt{\mu_sgr}. It does not depend on the car’s mass.

09
Front

What determines the ideal banking angle?

Back

For an ideal banked curve without friction, tan⁡θ=v2rg\tan\theta=\frac{v^2}{rg}, so θ=tan⁡−1(v2rg)\theta=\tan^{-1}\left(\frac{v^2}{rg}\right).

10
Front

Which part of tension turns a conical pendulum?

Back

In a conical pendulum, Tcos⁡θ=mgT\cos\theta=mg vertically and Tsin⁡θT\sin\theta supplies the inward radial force.

11
Front

How do radial equations differ at the top and bottom of a loop?

Back

At the top, both forces can point inward, so N+mg=mv2rN+mg=m\frac{v^2}{r}. At the bottom, gravity points outward, so N−mg=mv2rN-mg=m\frac{v^2}{r}.

12
Front

What is Newton’s law of universal gravitation?

Back

Newton’s law of universal gravitation is Fg=Gm1m2r2F_g=G\frac{m_1m_2}{r^2}. The force is attractive and acts along the line between the centers of mass.