Question
Problem 5: A 55.0-kg ice skater is moving at 4.00 m/s when she
grabs the loose end of a rope, the opposite end of which is tied to a
pole. She then moves in a circle of radius 0.800 m around the pole. (a)
Determine the force exerted by the horizontal rope on her arms.
(b) Compare this force with her weight.
Question 5:If the radius of the circle were doubled to 1.60 m while
keeping the skater’s speed constant at 4.00 m/s, what would be the new
tension in the rope?
Exercise 5:
A 55.0-kg ice skater is moving at 4.00 m/s when she grabs a rope tied to
a pole. If the radius of the circle she travels in is 0.800 m:
(a) Calculate the Force in the rope.
(b) How much greater is the force compared to the skater's weight?
Answers
GPT-4o mini
To solve the problem, we'll follow the steps below for parts (a) and (b), and then calculate the tension in the rope if the radius is doubled.
### **Given:**
- Mass of the skater, \( m = 55.0 \, \text{kg} \)
- Speed of the skater, \( v = 4.00 \, \text{m/s} \)
- Radius of the circle, \( r = 0.800 \, \text{m} \)
### **(a) Determine the force exerted by the horizontal rope on her arms:**
1. **Centripetal Force Calculation:**
The skater, while moving in a circular path, requires a centripetal force that is directed toward the center of the circle. The formula for centripetal force (\( F_c \)) is given by:
\[
F_c = \frac{mv^2}{r}
\]
Substituting in the values:
\[
F_c = \frac{55.0 \, \text{kg} \cdot (4.00 \, \text{m/s})^2}{0.800 \, \text{m}}
\]
\[
F_c = \frac{55.0 \, \text{kg} \cdot 16.00 \, \text{m}^2/\text{s}^2}{0.800 \, \text{m}}
\]
\[
F_c = \frac{880.0 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{0.800 \, \text{m}} = 1100 \, \text{N}
\]
Thus, the force exerted by the horizontal rope on her arms is **1100 N**.
### **(b) Compare this force with her weight:**
1. **Skater's Weight Calculation:**
The weight (\( W \)) of the skater is given by:
\[
W = mg
\]
where \( g = 9.81 \, \text{m/s}^2 \).
Substituting in the values:
\[
W = 55.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 539.55 \, \text{N}
\]
2. **Comparison:**
Now, we can compare the tension in the rope (which is the centripetal force) to her weight:
\[
\text{Comparison} = \frac{F_c}{W} = \frac{1100 \, \text{N}}{539.55 \, \text{N}} \approx 2.04
\]
Therefore, the force exerted by the rope is approximately **2.04 times greater** than her weight.
### **New Tension in the Rope when Radius is Doubled:**
If the radius of the circle is doubled to \( r = 1.60 \, \text{m} \):
1. **New Centripetal Force Calculation:**
Again using the formula for centripetal force:
\[
F_c' = \frac{mv^2}{r'}
\]
where \( r' = 1.60 \, \text{m} \).
Substituting in the values:
\[
F_c' = \frac{55.0 \, \text{kg} \cdot (4.00 \, \text{m/s})^2}{1.60 \, \text{m}}
\]
\[
F_c' = \frac{55.0 \cdot 16.00}{1.60} = \frac{880.0}{1.60} = 550 \, \text{N}
\]
Thus, when the radius is doubled, the new tension in the rope is **550 N**.
### **Given:**
- Mass of the skater, \( m = 55.0 \, \text{kg} \)
- Speed of the skater, \( v = 4.00 \, \text{m/s} \)
- Radius of the circle, \( r = 0.800 \, \text{m} \)
### **(a) Determine the force exerted by the horizontal rope on her arms:**
1. **Centripetal Force Calculation:**
The skater, while moving in a circular path, requires a centripetal force that is directed toward the center of the circle. The formula for centripetal force (\( F_c \)) is given by:
\[
F_c = \frac{mv^2}{r}
\]
Substituting in the values:
\[
F_c = \frac{55.0 \, \text{kg} \cdot (4.00 \, \text{m/s})^2}{0.800 \, \text{m}}
\]
\[
F_c = \frac{55.0 \, \text{kg} \cdot 16.00 \, \text{m}^2/\text{s}^2}{0.800 \, \text{m}}
\]
\[
F_c = \frac{880.0 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{0.800 \, \text{m}} = 1100 \, \text{N}
\]
Thus, the force exerted by the horizontal rope on her arms is **1100 N**.
### **(b) Compare this force with her weight:**
1. **Skater's Weight Calculation:**
The weight (\( W \)) of the skater is given by:
\[
W = mg
\]
where \( g = 9.81 \, \text{m/s}^2 \).
Substituting in the values:
\[
W = 55.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 539.55 \, \text{N}
\]
2. **Comparison:**
Now, we can compare the tension in the rope (which is the centripetal force) to her weight:
\[
\text{Comparison} = \frac{F_c}{W} = \frac{1100 \, \text{N}}{539.55 \, \text{N}} \approx 2.04
\]
Therefore, the force exerted by the rope is approximately **2.04 times greater** than her weight.
### **New Tension in the Rope when Radius is Doubled:**
If the radius of the circle is doubled to \( r = 1.60 \, \text{m} \):
1. **New Centripetal Force Calculation:**
Again using the formula for centripetal force:
\[
F_c' = \frac{mv^2}{r'}
\]
where \( r' = 1.60 \, \text{m} \).
Substituting in the values:
\[
F_c' = \frac{55.0 \, \text{kg} \cdot (4.00 \, \text{m/s})^2}{1.60 \, \text{m}}
\]
\[
F_c' = \frac{55.0 \cdot 16.00}{1.60} = \frac{880.0}{1.60} = 550 \, \text{N}
\]
Thus, when the radius is doubled, the new tension in the rope is **550 N**.