Question

chapter 4 - uniform circular motion & gravity

ex: a 65-kg skater travels at 2.0 m/s in a circle of radius 4.0 m.
a) what is her centripetal acceleration? b) what is the centripetal force? c) what actual force acts as the centripetal force? include a fbd in your response. show all work with units. (ans: a) \\(a_c = 1.0\text{ m/s}^2\\); b) \\(f_c = 65\text{ n}\\); c) friction between skater & ice surface)

ex: a 0.50 kg ball is tied to a string and swung in a horizontal circle of radius 1.2 m. the ball completes one full revolution in 0.80 s.
find: a) the centripetal acceleration & b) the centripetal force acting on the ball. c) what actual force provides the centripetal force? include a fbd in your response. show all work with units. (ans: a) \\(a_c = 74\text{ m/s}^2\\); b) \\(f_c = 37\text{ n}\\); c) tension)

ex: a 1.7-kg object is swung from the end of a 0.60 m string in a vertical circle. if the time of one revolution is 1.1 s, what is the tension at the top of the objects path? bottom of the objects path? include a fbd in your response. show all work with units. (ans: a) \\(f_t = 16.6\text{ n}\\); b) \\(f_t = 50\text{ n}\\))

Explanation:

Analyze the first example (Skater)

We are given:

• Mass of the skater, \(m = 65\text{ kg}\)
• Speed, \(v = 2.0\text{ m/s}\)
• Radius of the circular path, \(r = 4.0\text{ m}\)

We need to find:

• a) Centripetal acceleration, \(a_c\)
• b) Centripetal force, \(F_c\)
• c) The actual force acting as the centripetal force.

Let's calculate:

• a) Centripetal acceleration formula:

$$a_c = \frac{v^2}{r} = \frac{(2.0\text{ m/s})^2}{4.0\text{ m}} = \frac{4.0}{4.0} = 1.0\text{ m/s}^2$$

• b) Centripetal force formula:

$$F_c = m \cdot a_c = 65\text{ kg} \cdot 1.0\text{ m/s}^2 = 65\text{ N}$$

• c) The physical force that keeps the skater in a circular path on ice is the static friction force between the skater's blades and the ice surface.

Analyze the second example (Ball on a string in horizontal circle)

We are given:

• Mass of the ball, \(m = 0.50\text{ kg}\)
• Radius of the circle, \(r = 1.2\text{ m}\)
• Period of one full revolution, \(T = 0.80\text{ s}\)

We need to find:

• a) Centripetal acceleration, \(a_c\)
• b) Centripetal force, \(F_c\)
• c) The actual force providing the centripetal force.

Let's calculate:

• Speed \(v\) in terms of period \(T\):

$$v = \frac{2\pi r}{T} = \frac{2\pi (1.2\text{ m})}{0.80\text{ s}} = 3\pi\text{ m/s} \approx 9.425\text{ m/s}$$

• a) Centripetal acceleration:

$$a_c = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2} = \frac{4\pi^2 (1.2)}{(0.80)^2} = \frac{4.8\pi^2}{0.64} = 7.5\pi^2 \approx 74.02\text{ m/s}^2 \approx 74\text{ m/s}^2$$

• b) Centripetal force:

$$F_c = m \cdot a_c = 0.50\text{ kg} \cdot 74.02\text{ m/s}^2 \approx 37\text{ N}$$

• c) The actual force pulling the ball inward along the horizontal circle is the tension in the string.

Analyze the third example (Object in vertical circle)

We are given:

• Mass of the object, \(m = 1.7\text{ kg}\)
• Radius of the vertical circle, \(r = 0.60\text{ m}\)
• Period of one revolution, \(T = 1.1\text{ s}\)

We need to find:

• a) Tension at the top of the path, \(F_{T,\text{top}}\)
• b) Tension at the bottom of the path, \(F_{T,\text{bottom}}\)

Let's calculate:

• Centripetal acceleration:

$$a_c = \frac{4\pi^2 r}{T^2} = \frac{4\pi^2 (0.60)}{(1.1)^2} = \frac{2.4\pi^2}{1.21} \approx 19.57\text{ m/s}^2$$

• Gravity force acting on the object:

$$F_g = m \cdot g = 1.7\text{ kg} \cdot 9.8\text{ m/s}^2 = 16.66\text{ N}$$

• a) At the top of the path, both gravity and tension point downward towards the center:

$$F_{T,\text{top}} + F_g = F_c \implies F_{T,\text{top}} = m \cdot a_c - m \cdot g$$

$$F_{T,\text{top}} = 1.7\text{ kg} \cdot 19.57\text{ m/s}^2 - 16.66\text{ N} \approx 33.27\text{ N} - 16.66\text{ N} \approx 16.6\text{ N}$$

• b) At the bottom of the path, tension points upward (towards the center) and gravity points downward (away from the center):

$$F_{T,\text{bottom}} - F_g = F_c \implies F_{T,\text{bottom}} = m \cdot a_c + m \cdot g$$

$$F_{T,\text{bottom}} = 33.27\text{ N} + 16.66\text{ N} \approx 49.93\text{ N} \approx 50\text{ N}$$

Answer:

Question 1

Ex: A 65-kg skater travels at 2.0 m/s in a circle of radius 4.0 m.

• a) What is her centripetal acceleration?

$$a_c = \frac{v^2}{r} = \frac{(2.0\text{ m/s})^2}{4.0\text{ m}} = 1.0\text{ m/s}^2$$

• b) What is the centripetal force?

$$F_c = m \cdot a_c = 65\text{ kg} \cdot 1.0\text{ m/s}^2 = 65\text{ N}$$

• c) What actual force acts as the centripetal force?

The static friction force between the skater's blades and the ice surface.

---

Question 2

Ex: A 0.50 kg ball is tied to a string and swung in a horizontal circle of radius 1.2 m. The ball completes one full revolution in 0.80 s.

• a) Find the centripetal acceleration:

$$a_c = \frac{4\pi^2 r}{T^2} = \frac{4\pi^2 (1.2\text{ m})}{(0.80\text{ s})^2} \approx 74\text{ m/s}^2$$

• b) Find the centripetal force acting on the ball:

$$F_c = m \cdot a_c = 0.50\text{ kg} \cdot 74\text{ m/s}^2 \approx 37\text{ N}$$

• c) What actual force provides the centripetal force?

The tension force in the string.

---

Question 3

Ex: A 1.7-kg object is swung from the end of a 0.60 m string in a vertical circle. If the time of one revolution is 1.1 s, what is the tension at the top of the object's path? bottom of the object's path?

• a) Tension at the top of the path:

$$F_{T,\text{top}} = \frac{4\pi^2 m r}{T^2} - mg = \frac{4\pi^2 (1.7\text{ kg})(0.60\text{ m})}{(1.1\text{ s})^2} - (1.7\text{ kg})(9.8\text{ m/s}^2) \approx 16.6\text{ N}$$

• b) Tension at the bottom of the path:

$$F_{T,\text{bottom}} = \frac{4\pi^2 m r}{T^2} + mg = \frac{4\pi^2 (1.7\text{ kg})(0.60\text{ m})}{(1.1\text{ s})^2} + (1.7\text{ kg})(9.8\text{ m/s}^2) \approx 50\text{ N}$$