Related questions
Question
Transcribed Image Text:Estimate the moment of inertia of a 1.92 m tall, 107.2 kg heavy person's foot
segment (about its centre of mass).
Your Answer:
Answer
JO
kgm^2
units
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by stepSolved in 2 steps with 2 images
Knowledge Booster
Similar questions
- A 3 kg point mass is at coordinates (8 m, 7 m), a 7 kg mass is at (-9,3) and a 4kg mass is at (x,y) of (2,-4). Find the moment of inertia about the x axis. Ix = __________ kg-m2 ROUND TO 3 DECIMAL PLACESarrow_forwardProblem 3: Suppose we want to calculate the moment of inertia of a 65.5 kg skater, relative to a vertical axis through their center of mass. Part (a) First calculate the moment of inertia (in kg m2) when the skater has their arms pulled inward by assuming they are cylinder of radius 0.11 m. sin() cos() tan() 7 8 9 HOME cotan() asin() acos() E 4 5 atan() acotan() sinh() 1 3 cosh() tanh() cotanh() END ODegrees O Radians Vol BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Part (b) Now calculate the moment of inertia of the skater (in kg m?) with their arms extended by assuming that each arm is 5% of the mass of their body. Assume the body is a cylinder of the same size, and the arms are 0.825 m long rods extending straight out from the center of their body being rotated at the ends.arrow_forwardEveryone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of diameter Da = 0.273 m. The mass of the hoop part is mh = : 0.120 kg and the mass of the disk part is md = 0.050 kg. Imagine making a boomerang that has the same total moment of inertia around its center as the sport disk. The boomerang is to be constructed in the shape of an "X," which can be approximated as two thin, uniform rods joined at their midpoints. If the total mass of the boomerang is to be m₁ = 0.235 kg, what must be the length Lы of the boomerang? | I Lb = = Cross-sectional view marrow_forward
- Multiple-Concept Example 10 reviews the approach and some of the concepts that are pertinent to this problem. The figure shows a model for the motion of the human forearm in throwing a dart. Because of the force M applied by the triceps muscle, the forearm can rotate about an axis at the elbow joint. Assume that the forearm has the dimensions shown in the figure and a moment of inertia of 0.062 kg-m? (including the effect of the dart) relative to the axis at the elbow. Assume also that the force M acts perpendicular to the forearm. Ignoring the effect of gravity and any frictional forces, determine the magnitude of the force M needed to give the dart a tangential speed of 4.5 m/s in 0.14 s, starting from rest. 0.28 m Axis at elbow joint 0.025 m Number i Unitsarrow_forwardFour identical particles (mass of each = 0.40 kg) are placed at the vertices of a rectangle (2.0 m × 3.0 m) and held in those positions by four light rods which form the sides of the rectangle. What is the moment of inertia of this rigid body about an axis that passes through the mid-points of the longer sides and is parallel to the shorter sides? Group of answer choices 2.7 kg⋅m2 3.6 kg⋅m2 1.6 kg⋅m2 3.1 kg⋅m2 4.1 kg⋅m2arrow_forwardThe moment of inertia of the lower leg and foot about an axis through the knee joint is 0.23 kgm^2. What is the moment of inertia of the leg and foot about the knee joint if a 0.82 kg shoe is worn on the foot? Assume that the shoe's mass is all concentrated in one point 40 cm from the knee. (please report to two decimal places)arrow_forward
- The figure below shows a side view of a car tire before it is mounted on a wheel. Model it as having two sidewalls of uniform thickness 0.600 cm and a tread wall of uniform thickness 2.50 cm and width 19.8 cm. Assume the rubber has a uniform density 1.10 x 10³ kg/m³. Find its moment of inertia about an axis perpendicular to the page through its center. 0.082 x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kg - m² 33.0 cm 30.5 cm 16.5 cm Sidewall Treadarrow_forwardCalculate the moment of inertia of a skater given the following information. (a) The 92.0-kg skater is approximated as a cylinder that has a 0.120-m radius. kg · m2(b) The skater with arms extended is approximately a cylinder that is 86.0 kg, has a 0.120-m radius, and has two 0.850-m-long arms which are 3.00 kg each and extend straight out from the cylinder like rods rotated about their ends.arrow_forwardCalculate the moment of inertia (in kg·m2) of a skater given the following information. (a)The 88.0 kg skater is approximated as a cylinder that has a 0.130 m radius. ____kg·m2 (b)The skater with arms extended is approximately a cylinder that is 82.0 kg, has a 0.130 m radius, and has two 0.750 m long arms which are 3.00 kg each and extend straight out from the cylinder like rods rotated about their ends. _____kg·m2arrow_forward
- A uniform thin rod of mass M = 3.55 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m = 0.291 kg, are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I = 0.845 kg-m² ? L = m Larrow_forwardFind the moment of inertia about y-axis if each mass is equal to 1 kg (in kg×m2)arrow_forward
arrow_back_ios
arrow_forward_ios