Question

Direction:Think of situations or object which you can relate to uniform circular motion.use a phrase to explain it Fill in the blank bellow
1example: a racing car
2__________
3__________
4__________
5__________
6__________
7__________
8__________
9__________
10_________​

Answer 1

Explanation:

one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or

\[d\overset{\to }{v}\text{/}dt\ne 0.\]

This is shown in (Figure). As the particle moves counterclockwise in time

\[\text{Δ}t\]

on the circular path, its position vector moves from

\[\overset{\to }{r}(t)\]

to

\[\overset{\to }{r}(t+\text{Δ}t).\]

The velocity vector has constant magnitude and is tangent to the path as it changes from

\[\overset{\to }{v}(t)\]

to

\[\overset{\to }{v}(t+\text{Δ}t),\]

changing its direction only. Since the velocity vector

\[\overset{\to }{v}(t)\]

is perpendicular to the position vector

\[\overset{\to }{r}(t),\]

the triangles formed by the position vectors and

\[\text{Δ}\overset{\to }{r},\]

and the velocity vectors and

\[\text{Δ}\overset{\to }{v}\]

are similar. Furthermore, since

\[|\overset{\to }{r}(t)|=|\overset{\to }{r}(t+\text{Δ}t)|\]

and

\[|\overset{\to }{v}(t)|=|\overset{\to }{v}(t+\text{Δ}t)|,\]

the two triangles are isosceles. From these facts we can make the assertion

\[\frac{\text{Δ}v}{v}=\frac{\text{Δ}r}{r}\]

or

\[\text{Δ}v=\frac{v}{r}\text{Δ}r.\]

Figure a shows a circle with center at point C. We are shown radius r of t and radius r of t, which are an angle Delta theta apart, and the chord length delta r connecting the ends of the two radii. Vectors r of t, r of t plus delta t, and delta r form a triangle. At the tip of vector r of t, the velocity is shown as v of t and points up and to the right, tangent to the circle. . At the tip of vector r of t plus delta t, the velocity is shown as v of t plus delta t and points up and to the left, tangent to the circle. Figure b shows the vectors v of t and v of t plus delta t with their tails together, and the vector delta v from the tip of v of t to the tip of v of t plus delta t. These three vectors form a triangle. The angle between the v of t and v of t plus delta t is theta.

Figure 4.18 (a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times

\[t\]

and

\[t+\text{Δ}t.\]

(b) Velocity vectors forming a triangle. The two triangles in the figure are similar. The vector

\[\text{Δ}\overset{\to }{v}\]

points toward the center of the circle in the limit

\[\text{Δ}t\to 0.\]

We can find the magnitude of the acceleration from

\[a=\underset{\text{Δ}t\to 0}{\text{lim}}(\frac{\text{Δ}v}{\text{Δ}t})=\frac{v}{r}(\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\text{Δ}r}{\text{Δ}t})=\frac{{v}^{2}}{r}.\]

The direction of the acceleration can also be found by noting that as

\[\text{Δ}t\]

and therefore

\[\text{Δ}\theta\]

approach zero, the vector

\[\text{Δ}\overset{\to }{v}\]

approaches a direction perpendicular to

\[\overset{\to }{v}.\]

In the limit

\[\text{Δ}t\to 0,\]

\[\text{Δ}\overset{\to }{v}\]

is perpendicular to

\[\overset{\to }{v}.\]

Since

\[\overset{\to }{v}\]

is tangent to the circle, the acceleration

\[d\overset{\to }{v}\text{/}dt\]

points toward the center of the circle. Summarizing, a particle moving in a circle at a constant speed has an acceleration with magnitude

\[{a}_{\text{C}}=\frac{{v}^{2}}{r}.\]

The direction of the acceleration vector is toward the center of the circle ((Figure)). This is a radial acceleration and is called the centripetal acceleration, which is why we give it the subscript c. The word centripetal comes from the Latin words centrum (meaning “center”) and petere (meaning to seek”), and thus takes the meaning “center seeking.”

A circle is shown with a purple arrow labeled as vector a sub c pointing radially inward and a green arrow tangent to the circle and labeled v. The arrows are shown with their tails at the same point on the circle.

Answer 2

Question-

Direction: Think of situations or object which you can relate to uniform circular motion. Use a phrase to explain it Fill in the blank bellow

1 example: a racing car

2__________

3__________

4__________

5__________

6__________

7__________

8__________

9__________

10_________​

Answer-

1. Racing car
2. An artificial satellite orbiting the Earth at a constant height
3. A ceiling fan's blades rotating around a hub
4. A stone that is tied to a rope and is being swung in circles
5. A car turning through a curve in a race track
6. An electron moving perpendicular to a uniform magnetic field
7. A gear turning inside a mechanism.
8. The motion of a wheel.
9. The motion of the hands of a clock
10. The motion of the planets.