uniform circular motion and gravitation review
1. a 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. determine the acceleration and the net force acting upon the car.
2. a 95-kg halfback makes a turn on the football field. the halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. the halfback makes a quarter of a turn around the circle in 2.1 seconds. determine the speed, acceleration and net force acting upon the halfback
3. determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the cliffhanger in 29.3 seconds. the radius of the barrel is 2.90 meters.
4. the fact that satellites can maintain their motion and their distance above the earth is fascinating to many. how can it be? what keeps a satellite up?
5. if there is an inward force acting upon an earth orbiting satellite, then why doesnt the satellite collide into the earth?
6. a 2 kg ball on a string is rotated about a circle of radius 10 m. the maximum tension allowed in the string is 50 n. what is the maximum speed of the ball?

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Find centripetal acceleration
The formula for centripetal acceleration is $a_c = \frac{v^2}{r}$, where $v = 10\ m/s$ and $r = 25.0\ m$.
$a_c = \frac{(10)^2}{25.0} = \frac{100}{25.0} = 4\ m/s^2$

## Step2: Find net force (centripetal force)
The formula for centripetal force is $F_c = m \cdot a_c$, where $m = 900\ kg$ and $a_c = 4\ m/s^2$.
$F_c = 900 	imes 4 = 3600\ N$

# Answer:
Acceleration: $4\ m/s^2$, Net Force: $3600\ N$

### Problem 2
# Explanation:
## Step1: Find speed (v)
A quarter - turn means the distance traveled $d=\frac{1}{4}	imes2\pi r$. Here, $r = 12\ m$, so $d=\frac{1}{4}	imes2\pi	imes12 = 6\pi\ m$. Time $t = 2.1\ s$. Speed $v=\frac{d}{t}=\frac{6\pi}{2.1}\approx\frac{18.85}{2.1}\approx8.98\ m/s$

## Step2: Find centripetal acceleration
Using $a_c=\frac{v^2}{r}$, with $v\approx8.98\ m/s$ and $r = 12\ m$.
$a_c=\frac{(8.98)^2}{12}\approx\frac{80.64}{12}\approx6.72\ m/s^2$

## Step3: Find net force
Using $F_c = m\cdot a_c$, with $m = 95\ kg$ and $a_c\approx6.72\ m/s^2$.
$F_c=95	imes6.72 = 638.4\ N$

# Answer:
Speed: $\approx8.98\ m/s$, Acceleration: $\approx6.72\ m/s^2$, Net Force: $\approx638.4\ N$

### Problem 3
# Explanation:
## Step1: Find period (T)
Number of revolutions $n = 10$, time $t = 29.3\ s$. Period $T=\frac{t}{n}=\frac{29.3}{10}=2.93\ s$

## Step2: Find angular velocity ($\omega$)
$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.93}\approx2.14\ rad/s$

## Step3: Find centripetal acceleration
$a_c=\omega^2r$, where $r = 2.90\ m$.
$a_c=(2.14)^2	imes2.90\approx4.58	imes2.90\approx13.28\ m/s^2$

## Step4: Find centripetal force
$F_c = m\cdot a_c$, with $m = 40\ kg$.
$F_c=40	imes13.28 = 531.2\ N$

# Answer:
Centripetal Force: $\approx531.2\ N$

### Problem 4
# Brief Explanations:
Satellites maintain motion and distance because the gravitational force between the Earth and the satellite provides the centripetal force required for uniform circular motion. The satellite's tangential velocity and the inward gravitational pull balance such that the satellite falls towards the Earth but also moves forward, following a curved path (orbit) instead of colliding.

# Answer:
Gravitational force from Earth provides centripetal force for the satellite's orbital motion. The satellite's tangential velocity and gravitational pull balance, keeping it in orbit.

### Problem 5
# Brief Explanations:
The satellite doesn't collide because the centripetal force (inward gravitational force) causes it to follow a curved (orbital) path. The satellite has a tangential velocity, so as it falls towards the Earth due to gravity, it also moves forward, and the curvature of its path matches the curvature of the Earth, preventing a collision.

# Answer:
The satellite's tangential velocity and the centripetal (gravitational) force cause it to follow an orbital path, matching Earth's curvature, so it doesn't collide.

### Problem 6
# Explanation:
## Step1: Relate tension and centripetal force
The tension in the string provides the centripetal force, so $F_c = T_{max}=50\ N$. The formula for centripetal force is $F_c=\frac{mv^2}{r}$.

## Step2: Solve for v
We know $F_c = 50\ N$, $m = 2\ kg$, $r = 10\ m$. Rearranging $F_c=\frac{mv^2}{r}$ for $v$:
$v=\sqrt{\frac{F_c\cdot r}{m}}$
Substitute values: $v=\sqrt{\frac{50	imes10}{2}}=\sqrt{\frac{500}{2}}=\sqrt{250}\approx15.81\ m/s$

# Answer:
Maximum speed: $\approx15.81\ m/s$

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Find centripetal acceleration

Step2: Find net force (centripetal force)

Step1: Find speed (v)

Step2: Find centripetal acceleration

Step3: Find net force

Step1: Find period (T)

Step2: Find angular velocity ($\omega$)

Step3: Find centripetal acceleration

Step4: Find centripetal force

Answer:

Problem 2