**Title: Understanding the Balance of a Seesaw**The uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of 32.1 kg. What is the mass of his friend (in kg)?**Diagram Explanation:**- The seesaw is depicted as a long plank supported at its midpoint by a triangular fulcrum.- The left side shows a boy positioned 2.0 meters from the fulcrum.- The right side shows another boy positioned 4.0 meters from the fulcrum.- The distance between the fulcrum and each boy is indicated with arrows labeled with their respective distances: 2.0 meters for the boy on the left and 4.0 meters for the boy on the right.**Objective:**Calculate the mass of the boy on the left such that the seesaw remains balanced. Enter the answer in the designated box marked in kilograms. **Hint:**Use the principle of moments which states that for the seesaw to be balanced, the moments on either side of the fulcrum must be equal.\[ \text{Moment on the left} = \text{Moment on the right} \]Moment is calculated by multiplying the mass by the distance from the fulcrum. ### Educational Exercise on Angular Momentum#### Problem (a)Calculate the angular momentum (in kg·m²/s) of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.450 kg·m².\[ \text{Result:} \square \, \text{kg·m²/s} \]#### Problem (b)He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kg·m²) if his angular velocity drops to 1.00 rev/s.\[ \text{Result:} \square \, \text{kg·m²} \]#### Problem (c)Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque (in N·m) was exerted if this takes 18.0 seconds?\[ \text{Result:} \square \, \text{N·m} \]---### Notes:- The questions involve calculating physical quantities related to rotational motion, utilizing concepts like angular momentum, moment of inertia, angular velocity, and torque. - Each problem builds upon the previous, demonstrating how changes in physical parameters affect motion.- This exercise is aimed at students studying physics, specifically rotational dynamics.

Given data: The mass of smaller boy (m1) = 32.1 kg The distance of smaller boy from pivot (r1) = 4.0 m The distance of his friend from pivot (r2) = 2.0 m Required: Mass of his friend (m2)Since you have asked multiple question, as per our guidelines we are supposed to answer only one question. Kindly repost other questions as the next question. The representation of forces and torques g = acceleration due to gravity The torques due to weight of the boys on the seesaw are in opposition directions so to maintain the balance of seesaw, the magnitude of the torques must be equal. τ2 = τ1r2F2= r1F1r2m2g= r1m1gm2 =r1r2×m1= 4.0 m2.0 m×32.1 kg= 64.2 kg The mass of his friend is 64.2 kg Answer: 64.2

Answered: The uniform seesaw is balanced at its…

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The uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of 32.1 kg. What is the mass of his friend (in kg)? F2.0 m -2.0 m→- 4.0 m- kg

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Center of Gravity of a system

The centre of gravity is a property that is based on the geometry of a rigid body or any particle. It is described as an equilibrium point or a hypothetical location where the total weight of the body is concentrated.

Question

**Title: Understanding the Balance of a Seesaw**

The uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of 32.1 kg. What is the mass of his friend (in kg)?

**Diagram Explanation:**

- The seesaw is depicted as a long plank supported at its midpoint by a triangular fulcrum.
- The left side shows a boy positioned 2.0 meters from the fulcrum.
- The right side shows another boy positioned 4.0 meters from the fulcrum.
- The distance between the fulcrum and each boy is indicated with arrows labeled with their respective distances: 2.0 meters for the boy on the left and 4.0 meters for the boy on the right.

**Objective:**

Calculate the mass of the boy on the left such that the seesaw remains balanced. Enter the answer in the designated box marked in kilograms.

**Hint:**

Use the principle of moments which states that for the seesaw to be balanced, the moments on either side of the fulcrum must be equal.

\[ \text{Moment on the left} = \text{Moment on the right} \]

Moment is calculated by multiplying the mass by the distance from the fulcrum.

### Educational Exercise on Angular Momentum

#### Problem (a)
Calculate the angular momentum (in kg·m²/s) of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.450 kg·m².

\[ \text{Result:} \square \, \text{kg·m²/s} \]

#### Problem (b)
He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kg·m²) if his angular velocity drops to 1.00 rev/s.

\[ \text{Result:} \square \, \text{kg·m²} \]

#### Problem (c)
Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque (in N·m) was exerted if this takes 18.0 seconds?

\[ \text{Result:} \square \, \text{N·m} \]

---
### Notes:
- The questions involve calculating physical quantities related to rotational motion, utilizing concepts like angular momentum, moment of inertia, angular velocity, and torque.
- Each problem builds upon the previous, demonstrating how changes in physical parameters affect motion.
- This exercise is aimed at students studying physics, specifically rotational dynamics.

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