circuluar motioncentripetal forcenormal forcefrictionless

0504 Dynamics of Circular Motion

Important Idea: Any object undergoing circular motion will experience acceleration (centripetal acceleration a⃗c\vec a_c or a⃗r\vec a_r), and by Newton's Second Law, it takes a Net Force to "force" that obect to undergo circular motion. That force is the centripetal force (FcF_c or FrF_r), which is proportional to centripetal acceleration a⃗c\vec a_c.

F⃗c=ma⃗c=mv2R\vec F_c = m\vec a_c = \frac{mv^2}{R}

Exercises

(Mid Term 2) A ball of mass mm, at the end of a string of length LL, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. The speed of the ball at the bottom of the circle is?
Graph

Solution

Todo
5gR\sqrt {5gR}

(Mid Term 2) A ball of mass 8.0kg is suspended by two wired from a horizontal arm, which is attached to a vertical shaft, as show in graph. The shaft is in uniform rotation about its axis such that the linear speed of the ball equals 2.3m/s. The tension in wire 2 is closed to:
Graph

Solution

Todo
66N

(Mid Term 2) A small object slides along the frictionless loop-the-loop with a diameter of 3m. What minimum speed must it have at the top of the loop?

Solution

Todo
3.8m/s

(Mid Term 2) A ball of mass mm is suspended from a rope of length R. The ball is set into freely-swinging circular motion in a vertical plane. The centripetal acceleration of the ball at the top of the circle is 8g8g. The centripetal acceleration of the ball at the bottom of the circle is closed to

Solution

Todo
12g12g

1. A car is moving along an inclined frictionless circular track. The inclined angle with the horizontal line is θ=30°\th = 30\degree, and the radius of the track is 100100m. Find the velocity vv, so that the car does not move up or down.
Graph (1)

Solution

tan⁡θ=FcWtan⁡30°=mv2Rmg⇒v=gRtan⁡30°=9.8×100×12≈23.8m/s\begin{aligned} \tan \th &= \frac{F_c}{W}\\ \tan 30\degree &= \frac{\frac{mv^2}{R}}{mg}\\ \To v &= \sqrt {gR\tan 30\degree}\\ &= \sqrt {9.8 \times 100 \times \frac{1}{2}} \approx 23.8m/s \end{aligned}

2. A car is at the bottom of a circular track with radius of 1010m, and moves along the track.
a. Find the initial velocity v0v_0, so that it just makes over the top.
Graph (2a)

Solution

At the top of the track, the centripetal force is provided solely by the car's weight and the normal force is zero. Let the centripetal force be FcF_c, the weight of the car be ww, and the normal be FNF_N.
1. According to Newton's Second Law

Fc=mvf2R=w−FNw=mgFN=0⇒vf=gR\begin{aligned} F_c &= \frac{mv_f^2}{R} = w - F_N\\ w &= mg\\ F_N &= 0\\ \To v_f &= \sqrt {gR} \end{aligned}

2. By Conservation of Energy

12mv02=12mvf2+mg⋅2R12mv02=12m⋅(gR)2+mg⋅2R⇒v0=5gR≈22.14m/s\begin{aligned} \frac{1}{2}mv_0^2 &= \frac{1}{2}mv_f^2 + mg\cdot 2R\\ \frac{1}{2}mv_0^2 &= \frac{1}{2}m \cdot (\sqrt{gR})^2 + mg\cdot 2R\\ \To v_0 &= \sqrt {5gR} \approx 22.14m/s \end{aligned}

b. At what height above ground does car leave track, if v0v_0 is 20% less than the initial velocity from part (a).
Graph (2b)

Solution

In this case, v0=0.8gRv_0 = 0.8\sqrt {gR}.
Whent the car leaves the track, the normal force is zero.

Fc=mvf2R=wcos⁡θ−FNw=mgFN=0⇒vf=gRcos⁡θ\begin{aligned} F_c &= \frac{mv_f^2}{R} = w\cos \th - F_N\\ w &= mg\\ F_N &= 0\\ \To v_f &= \sqrt {gR \cos \th} \end{aligned}

2. By Conservation of Energy, the height of the car is yf=R(1+cos⁡θ)y_f = R(1+\cos \th)

12mv02=12mvf2+mg⋅yf12m(0.8gR)2=12m⋅(gRcos⁡θ)2+mgR(1+cos⁡θ)⇒cos⁡θ=0.4yf=R(1+cos⁡θ)=10⋅(1+0.4)=14m\begin{aligned} \frac{1}{2}mv_0^2 &= \frac{1}{2}mv_f^2 + mg\cdot y_f\\ \frac{1}{2}m(0.8\sqrt{gR})^2 &= \frac{1}{2}m \cdot (\sqrt{gR \cos \th})^2 + mgR(1+\cos \th)\\ \To \cos \th &= 0.4\\ y_f &= R(1+\cos \th) &\\ &= 10\cdot (1+0.4) = 14m \end{aligned}

The height of the car is 1414m.