Math Problem Statement

A thin, cylindrical rod 36 cm long has mass 1.91 kg and radius 1.04 cm. A 29.2 kg ball of diameter 8.03 cm is attached to one end.

The arrangement is originally vertical with the ball at the top and is free to pivot about the other end.

8.03 cm

C

kg

1.91

90°

29.2 kg

pivot .

1.91 kg

O

• - 36 cm - »

After the ball-rod system falls a quarter turn, what is its rotational kinetic energy?

The acceleration of gravity is 9.8 m/s?.

Answer in units of J. Answer in units of J.

part 2 of 3

At the same point, what is its angular speed?

of rad/s.

Answer in units of rads. Answer in units

part 3 of 3

At the same point, what is the linear speed of the ball?

ms.

Answer in units of m/s. Answer in units of

Solution

To solve this problem, we need to analyze the system as it pivots from the vertical position to a quarter turn (90 degrees). We'll use the principles of rotational dynamics and energy conservation.

Part 1: Rotational Kinetic Energy (KE)

  1. Calculate the Moment of Inertia (I):

    • For the rod (cylinder), the moment of inertia about one end is given by: Irod=13mrodL2I_{\text{rod}} = \frac{1}{3} m_{\text{rod}} L^2 where mrod=1.91kgm_{\text{rod}} = 1.91 \, \text{kg} and L=0.36mL = 0.36 \, \text{m}.

    • For the ball (treated as a point mass at distance rr from the pivot): Iball=mballr2I_{\text{ball}} = m_{\text{ball}} r^2 where mball=29.2kgm_{\text{ball}} = 29.2 \, \text{kg} and r=0.36m+0.04m=0.40mr = 0.36 \, \text{m} + 0.04 \, \text{m} = 0.40 \, \text{m} (adding the radius of the ball).

  2. Calculate I for both components:

    • For the rod: Irod=13(1.91)(0.36)2=1.910.129630.082kg m2I_{\text{rod}} = \frac{1}{3} (1.91) (0.36)^2 = \frac{1.91 \cdot 0.1296}{3} \approx 0.082 \, \text{kg m}^2
    • For the ball: Iball=29.2(0.40)2=29.20.16=4.672kg m2I_{\text{ball}} = 29.2 \cdot (0.40)^2 = 29.2 \cdot 0.16 = 4.672 \, \text{kg m}^2
  3. Total Moment of Inertia (I_total): Itotal=Irod+Iball0.082+4.6724.754kg m2I_{\text{total}} = I_{\text{rod}} + I_{\text{ball}} \approx 0.082 + 4.672 \approx 4.754 \, \text{kg m}^2

  4. Calculate the gravitational potential energy (PE) before the fall:

    • The height of the center of mass of the ball when vertical is: hball=0.40mh_{\text{ball}} = 0.40 \, \text{m}
    • PE at the top: PE=mballgh=29.29.80.40114.56JPE = m_{\text{ball}} g h = 29.2 \cdot 9.8 \cdot 0.40 \approx 114.56 \, \text{J}
  5. At the bottom (after falling 90 degrees), all potential energy converts to kinetic energy: KE=PE114.56JKE = PE \approx 114.56 \, \text{J}

Part 2: Angular Speed (ω)

  1. Use the relationship: KE=12Itotalω2KE = \frac{1}{2} I_{\text{total}} \omega^2

  2. Solve for ω: 114.56=124.754ω2114.56 = \frac{1}{2} \cdot 4.754 \cdot \omega^2 ω2=229.124.75448.21\omega^2 = \frac{229.12}{4.754} \approx 48.21 ω48.216.95rad/s\omega \approx \sqrt{48.21} \approx 6.95 \, \text{rad/s}

Part 3: Linear Speed (v) of the Ball

  1. Use the relation: v=rωv = r \cdot \omega where r=0.40mr = 0.40 \, \text{m}: v=0.406.952.78m/sv = 0.40 \cdot 6.95 \approx 2.78 \, \text{m/s}

Final Answers:

  1. Rotational Kinetic Energy: 114.56J\approx 114.56 \, \text{J}
  2. Angular Speed: 6.95rad/s\approx 6.95 \, \text{rad/s}
  3. Linear Speed of the Ball: 2.78m/s\approx 2.78 \, \text{m/s}

If you have any questions or need further details, let me know! Here are some related questions to consider:

  1. What would happen to the kinetic energy if the mass of the ball increased?
  2. How does the radius of the rod affect the moment of inertia?
  3. What would be the potential energy at different angles of the pivot?
  4. How would friction at the pivot affect the system's energy?
  5. Can you derive the angular speed using conservation of energy principles?

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Math Problem Analysis

Mathematical Concepts

Rotational Dynamics
Energy Conservation
Kinematics

Formulas

I_rod = (1/3) * m_rod * L^2
I_ball = m_ball * r^2
KE = PE
KE = (1/2) * I_total * ω^2
v = r * ω

Theorems

Conservation of Energy
Moment of Inertia

Suitable Grade Level

Grades 11-12