Question
A 1.5 kg metal object is attached to the end of a string and whirled around in a circular
path of radius 6 m. If the stone completes 5 revolutions in 12 s, determine its;
(a) linear velocity during this period, and
(4)
(b) centripetal force.
path of radius 6 m. If the stone completes 5 revolutions in 12 s, determine its;
(a) linear velocity during this period, and
(4)
(b) centripetal force.
Answers
GPT-4o mini
To solve this problem, we'll start by determining the linear velocity of the metal object and then calculate the centripetal force acting on it.
### (a) Linear Velocity
Linear velocity (\(v\)) can be calculated using the formula:
\[
v = \frac{d}{t}
\]
where \(d\) is the distance traveled and \(t\) is the time taken.
First, we need to calculate the distance traveled in one revolution. The circumference (\(C\)) of a circle is given by the formula:
\[
C = 2 \pi r
\]
where \(r\) is the radius of the circular path.
Given that \(r = 6\, \text{m}\):
\[
C = 2 \pi (6) = 12\pi \, \text{m}
\]
Now, if the object completes 5 revolutions, the total distance (\(d\)) traveled is:
\[
d = 5 \cdot C = 5 \cdot 12\pi = 60\pi \, \text{m}
\]
Now, given that the time taken (\(t\)) is 12 seconds, we can calculate the linear velocity:
\[
v = \frac{d}{t} = \frac{60\pi}{12} = 5\pi \, \text{m/s}
\]
### (b) Centripetal Force
The centripetal force (\(F_c\)) acting on an object moving in a circle can be calculated using the formula:
\[
F_c = \frac{m v^2}{r}
\]
where \(m\) is the mass, \(v\) is the linear velocity, and \(r\) is the radius of the circular path.
Given:
- Mass (\(m\)) = 1.5 kg
- Linear velocity (\(v\)) = \(5\pi \, \text{m/s}\)
- Radius (\(r\)) = 6 m
First, we need to calculate \(v^2\):
\[
v^2 = (5\pi)^2 = 25\pi^2
\]
Now substituting the values into the centripetal force formula:
\[
F_c = \frac{1.5 \cdot 25\pi^2}{6}
\]
\[
F_c = \frac{37.5\pi^2}{6} = 6.25\pi^2 \, \text{N}
\]
To calculate the numeric value (using \(\pi \approx 3.14\)):
\[
F_c \approx 6.25 \cdot (3.14)^2 \approx 6.25 \cdot 9.8596 \approx 61.66 \, \text{N}
\]
### Final Answers
(a) Linear Velocity: \(5\pi \, \text{m/s} \approx 15.71 \, \text{m/s}\)
(b) Centripetal Force: \(6.25\pi^2 \, \text{N} \approx 61.66 \, \text{N}\)
### (a) Linear Velocity
Linear velocity (\(v\)) can be calculated using the formula:
\[
v = \frac{d}{t}
\]
where \(d\) is the distance traveled and \(t\) is the time taken.
First, we need to calculate the distance traveled in one revolution. The circumference (\(C\)) of a circle is given by the formula:
\[
C = 2 \pi r
\]
where \(r\) is the radius of the circular path.
Given that \(r = 6\, \text{m}\):
\[
C = 2 \pi (6) = 12\pi \, \text{m}
\]
Now, if the object completes 5 revolutions, the total distance (\(d\)) traveled is:
\[
d = 5 \cdot C = 5 \cdot 12\pi = 60\pi \, \text{m}
\]
Now, given that the time taken (\(t\)) is 12 seconds, we can calculate the linear velocity:
\[
v = \frac{d}{t} = \frac{60\pi}{12} = 5\pi \, \text{m/s}
\]
### (b) Centripetal Force
The centripetal force (\(F_c\)) acting on an object moving in a circle can be calculated using the formula:
\[
F_c = \frac{m v^2}{r}
\]
where \(m\) is the mass, \(v\) is the linear velocity, and \(r\) is the radius of the circular path.
Given:
- Mass (\(m\)) = 1.5 kg
- Linear velocity (\(v\)) = \(5\pi \, \text{m/s}\)
- Radius (\(r\)) = 6 m
First, we need to calculate \(v^2\):
\[
v^2 = (5\pi)^2 = 25\pi^2
\]
Now substituting the values into the centripetal force formula:
\[
F_c = \frac{1.5 \cdot 25\pi^2}{6}
\]
\[
F_c = \frac{37.5\pi^2}{6} = 6.25\pi^2 \, \text{N}
\]
To calculate the numeric value (using \(\pi \approx 3.14\)):
\[
F_c \approx 6.25 \cdot (3.14)^2 \approx 6.25 \cdot 9.8596 \approx 61.66 \, \text{N}
\]
### Final Answers
(a) Linear Velocity: \(5\pi \, \text{m/s} \approx 15.71 \, \text{m/s}\)
(b) Centripetal Force: \(6.25\pi^2 \, \text{N} \approx 61.66 \, \text{N}\)