Let's break down each question and in **two steps** with explanations: --- ## **1. of Inertia Calculation** **Step 1:** Recall the formula for the moment inertia of a rod about its center: \[ I = \frac{1}{12} m L^2] where \( m \) is mass and \( \) is length. - **arm:** \( = 1) kg, \( L 30 \) cm .3 m [ I_{\text{forearm}} = \frac{1}{12} \times 1 \times (.3)^2 = \frac{1}{12} \times .09 = .0075 \text{ kg} \cdot \text{m}^2 \] - **Hand:** \( m = .5 \) kg, \( L = 10 \) cm = .1 m \[ I_{\text{hand}} = \frac{1}{12} \times .5 \times (.1)^2 = \frac{1}{12} \times .5 \times .01 = .00042 \text{ kg} \cdot \text{m}^2 \] **Explanation:** The moment of inertia gives a measure of how much torque is needed for a desired angular acceleration about the center of mass. --- ## **2. Total External Torque with 5 kg Dumbbell** **Step 1:** Calculate the torque due to the dumbbell about the elbow joint: \[\text{Torque} = \text{Force} \times \text{distance from axis}\] The dumbbell is at the hand's center of mass, which is 30 cm (forearm) + 7 cm (hand's center from wrist) = **37 cm = .37 m** from the elbow. \[\text{Force (weight)} = 5 \text{ kg} \times 9.81 \text{ m/s}^2 = 49.05 \text{ N}\] \[\text{Torque}_\text{dumbbell} = 49.05 \times .37 = 18.15 \text{ Nm}\] **Explanation:** The torque is the product of the weight and its perpendicular distance from the axis of rotation (elbow). --- ## **3. Bicep Muscle Force for Equilibrium at 90°** **Step 1:** Sum all torques about the elbow (taking counterclockwise as positive): - **Dumbbell:** \( 18.15 \) Nm (clockwise) - **Forearm weight:** \[ F = 1 \text{ kg} \times 9.81 = 9.81 \text{ N} \] at 15 cm = .15 m \[ \tau_{\text{forearm}} = 9.81 \times .15 = 1.47 \text{ Nm} \] (clockwise) - **Hand weight:** \[ F = .5 \times 9.81 = 4.905 \text{ N} \] at 30 cm + 7 cm = 37 cm = .37 m \[ \tau_{\text{hand}} = 4.905 \times .37 = 1.815 \text{ Nm} \] (clockwise) Total clockwise torque: \[ 18.15 + 1.47 + 1.815 = 21.435 \text{ Nm} \] **Step 2:** The bicep force (\( F_b \)) acts at 4.5 cm = .045 m from the elbow, perpendicular. Set counterclockwise torque = clockwise torque for equilibrium: \[ F_b \times .045 = 21.435 \] \[ F_b = \frac{21.435}{.045} = 476.33 \text{ N} \] **Explanation:** The bicep must exert a large force close to the elbow because the moment arm is short. --- ## **4. Net Torque at 90° Position** **Step 1:** Sum all torques (from previous step): - Clockwise: \( 21.435 \) Nm - Counterclockwise (bicep): \( 476.33 \times .045 = 21.435 \) Nm Net torque: \[ \text{Net Torque} = 21.435 - 21.435 = \text{ Nm} \] **Step 2:** Since net torque = , the forearm **is in rotational equilibrium**. **Explanation:** Rotational equilibrium occurs when the sum of clockwise and counterclockwise torques is zero. --- ## **5. Bicep Force at 45° Position** **Step 1:** Now the bicep attaches 3.25 cm = .0325 m from the elbow. The perpendicular distance for each weight is multiplied by \( \sin(45^\circ) = .707 \): - Dumbbell: \( .37 \times .707 = .2616 \) m - Forearm: \( .15 \times .707 = .106 \) m - Hand: \( .37 \times .707 = .2616 \) m New torques: - Dumbbell: \( 49.05 \times .2616 = 12.83 \) Nm - Forearm: \( 9.81 \times .106 = 1.04 \) Nm - Hand: \( 4.905 \times .2616 = 1.284 \) Nm Total clockwise: \( 12.83 + 1.04 + 1.284 = 15.154 \) Nm **Step 2:** For equilibrium, \[ F_b \times .0325 = 15.154 \] \[ F_b = \frac{15.154}{.0325} = 466.28 \text{ N} \] **Explanation:** With a shorter bicep moment arm and reduced perpendicular distances, the force requirement remains high. --- ## **Final Answers:** 1. **Moment of inertia:** - Forearm: \( .0075 \text{ kg} \cdot \text{m}^2 \) - Hand: \( .00042 \text{ kg} \cdot \text{m}^2 \) 2. **Total external torque (dumbbell):** - \( 18.15 \) Nm 3. **Bicep force at 90°:** - \( 476.33 \) N 4. **Net torque at 90°:** - \( \) Nm (**Rotational equilibrium achieved**) 5. **Bicep force at 45°:** - \( 466.28 \) N

# Biomechanics Analysis of a Bicep Curl Exercise ## **1. Moment of Inertia Calculation** ### Step 1: Calculate Moment of Inertia Using the formula for the moment of inertia of a rod about its center: \[ I = \frac{1}{12} m L^2 \] - **Forearm:** - Mass (\(m\)) = 1 kg - Length (\(L\)) = 30 cm = 0.3 m \[ I_{\text{forearm}} = \frac{1}{12} \times 1 \times (0.3)^2 = \frac{1}{12} \times 0.09 = 0.0075 \text{ kg} \cdot \text{m}^2 \] - **Hand:** - Mass (\(m\)) = 0.5 kg - Length (\(L\)) = 10 cm = 0.1 m \[ I_{\text{hand}} = \frac{1}{12} \times 0.5 \times (0.1)^2 = \frac{1}{12} \times 0.5 \times 0.01 = 0.00042 \text{ kg} \cdot \text{m}^2 \] ### Explanation The moment of inertia shows how much torque is required to achieve a specific angular acceleration around the center of mass. --- ## **2. Total External Torque with 5 kg Dumbbell** ### Step 1: Calculate Torque from Dumbbell Using the torque formula: \[ \text{Torque} = \text{Force} \times \text{distance from axis} \] - **Force (Weight of Dumbbell):** \[ \text{Force} = 5 \text{ kg} \times 9.81 \text{ m/s}^2 = 49.05 \text{ N} \] - **Distance from Elbow:** Total distance = length of forearm + distance of hand's center of mass = 30 cm + 7 cm = 37 cm = 0.37 m \[ \text{Torque}_{\text{dumbbell}} = 49.05 \times 0.37 = 18.15 \text{ Nm} \] ### Explanation The torque produced by the dumbbell is the product of its weight and the distance from the elbow joint, indicating the moment it creates. --- ## **3. Bicep Muscle Force for Equilibrium at 90°** ### Step 1: Calculate Total Clockwise Torque Total clockwise torque comprises the contributions from the dumbbell, forearm, and hand: - **Forearm Weight:** \[ F = 1 \text{ kg} \times 9.81 \text{ m/s}^2 = 9.81 \text{ N} \quad \text{(at 15 cm = 0.15 m)} \] \[ \tau_{\text{forearm}} = 9.81 \times 0.15 = 1.47 \text{ Nm} \] - **Hand Weight:** \[ F = 0.5 \text{ kg} \times 9.81 \text{ m/s}^2 = 4.905 \text{ N} \quad \text{(at 37 cm = 0.37 m)} \] \[ \tau_{\text{hand}} = 4.905 \times 0.37 = 1.815 \text{ Nm} \] - **Total Clockwise Torque:** \[ \text{Total Clockwise Torque} = 18.15 + 1.47 + 1.815 = 21.435 \text{ Nm} \] ### Step 2: Calculate Bicep Muscle Force The bicep force (\(F_b\)) acts at a distance of 4.5 cm = 0.045 m from the elbow: \[ F_b \times 0.045 = 21.435 \] \[ F_b = \frac{21.435}{0.045} = 476.33 \text{ N} \] ### Explanation The bicep force must counteract the total torque produced by the weights to maintain equilibrium, highlighting the mechanical advantage and leverage involved. --- ## **4. Net Torque at 90° Position** ### Step 1: Calculate Net Torque From the previous step, the total clockwise torque is: - Clockwise Torque: \( 21.435 \) Nm - Counterclockwise Torque (bicep): \[ \tau_{\text{bicep}} = 476.33 \times 0.045 = 21.435 \text{ Nm} \] ### Step 2: Determine Equilibrium Net torque: \[ \text{Net Torque} = \text{Counterclockwise} - \text{Clockwise} = 21.435 - 21.435 = 0 \text{ Nm} \] ### Explanation Since the net torque is zero, the forearm is in rotational equilibrium at this position. --- ## **5. Bicep Force at 45° Position** ### Step 1: Recalculate Torque at 45° Angle The attachment point for the bicep is now at 3.25 cm = 0.0325 m. The distances are multiplied by \(\sin(45^\circ) = 0.707\): - **Dumbbell Torque:** \[ \text{Distance} = 0.37 \times 0.707 = 0.2616 \text{ m} \] \[ \tau_{\text{dumbbell}} = 49.05 \times 0.2616 = 12.83 \text{ Nm} \] - **Forearm Torque:** \[ \text{Distance} = 0.15 \times 0.707 = 0.106 \text{ m} \] \[ \tau_{\text{forearm}} = 9.81 \times 0.106 = 1.04 \text{ Nm} \] - **Hand Torque:** \[ \tau_{\text{hand}} = 4.905 \times 0.2616 = 1.284 \text{ Nm} \] - **Total Clockwise Torque:** \[ \text{Total Clockwise Torque} = 12.83 + 1.04 + 1.284 = 15.154 \text{ Nm} \] ### Step 2: Calculate New Bicep Force For equilibrium: \[ F_b \times 0.0325 = 15.154 \] \[ F_b = \frac{15.154}{0.0325} = 466.28 \text{ N} \] ### Explanation At a 45° angle, the required bicep force decreases slightly but remains significant due to the weight and distances involved. --- ## **Final Answers:** 1. **Moment of Inertia:** - Forearm: \( 0.0075 \text{ kg} \cdot \text{m}^2 \) - Hand: \( 0.00042 \text{ kg} \cdot \text{m}^2 \) 2. **Total External Torque (Dumbbell):** - \( 18.15 \) Nm 3. **Bicep Force at 90°:** - \( 476.33 \) N 4. **Net Torque at 90°:** - \( 0 \) Nm (Rotational equilibrium achieved) 5. **Bicep Force at 45°:** - \( 466.28 \) N