The person whose leg is shown in Figure 10.22 kicks his leg by exerting a 2000-N force with his upper legmuscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is1.25 kg - m2, (a) find the angular acceleration of the leg. (b) Neglecting the gravitational force, what is therotational kinetic energy of the leg after it has rotated through 57.3° (1.00 rad)?Fr. = 2.20 cmFemur+\= 1.25 kg • m2Figure 10.22 The muscle in the upper leggives the lower leg an angular accelerationand imparts rotational kinetic energy to it byexerting a torque about the knee. Fis a vectorthat is perpendicular to r. This exampleexamines the situation.Solution to (a)From the rotational analog to Newton's second law, the angular acceleration a isnet Ta =10.101Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight doesnot create a torque, the net torque is thusnet T(0.0220 m)(2000 N)= 44.0 N · m.10.102Substituting this value for the torque and the given value for the moment of inertia into the expression fora gives44.0 N · m1.25 kg · m2= 35.2 rad/s².a =10.103 3.3. In the OpenStax College Physics textbook, read the Chapter 10 sections 10.1 (AngularAcceleration), 10.3 (Dynamics of Rotational Motion: Rotational Inertia) and 10.5 (AngularMomentum and Its Conservation). Then write answers to [a] and [b] below.For all your answers, please write down your calculations and explain your reasoning. Drawsketches whenever helpful, and please use your units correctly. If you are using any onlineresources (such as unit conversion tools) give their URL and webpage title in your answers.[a] For Example 10.13 (Calculating the Torque in a Kick) in Section 10.5, replace “2000-N force"with "X-N force" where X is the number you computed on the first page of this study guide.Then solve part (a) of the example. There is no need to solve part (b).X = 883%3D

First we need to find torque due to force X-N and then using moment of inertia we can calculate…

[a] For Example 10.13 (Calculating the Torque in a Kick) in Section 10.5, replace "2000-N force" with "X-N force" where X is the number you computed on the first page of this study guide. Then solve part (a) of the example. There is no need to solve part (b). X = 883

[a] For Example 10.13 (Calculating the Torque in a Kick) in Section 10.5, replace "2000-N force" with "X-N force" where X is the number you computed on the first page of this study guide. Then solve part (a) of the example. There is no need to solve part (b). X = 883

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter11: Angular Momentum

Section: Chapter Questions

Problem 54P: A cylinder with rotational inertia I1=2.0kgm2 rotates clockwise about a vertical axis through its...

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Concept explainers

Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

Question

The person whose leg is shown in Figure 10.22 kicks his leg by exerting a 2000-N force with his upper leg
muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is
1.25 kg - m2, (a) find the angular acceleration of the leg. (b) Neglecting the gravitational force, what is the
rotational kinetic energy of the leg after it has rotated through 57.3° (1.00 rad)?
F
r. = 2.20 cm
Femur
+\= 1.25 kg • m2
Figure 10.22 The muscle in the upper leg
gives the lower leg an angular acceleration
and imparts rotational kinetic energy to it by
exerting a torque about the knee. Fis a vector
that is perpendicular to r. This example
examines the situation.
Solution to (a)
From the rotational analog to Newton's second law, the angular acceleration a is
net T
a =
10.101
Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does
not create a torque, the net torque is thus
net T
(0.0220 m)(2000 N)
= 44.0 N · m.
10.102
Substituting this value for the torque and the given value for the moment of inertia into the expression for
a gives
44.0 N · m
1.25 kg · m2
= 35.2 rad/s².
a =
10.103

3.3. In the OpenStax College Physics textbook, read the Chapter 10 sections 10.1 (Angular
Acceleration), 10.3 (Dynamics of Rotational Motion: Rotational Inertia) and 10.5 (Angular
Momentum and Its Conservation). Then write answers to [a] and [b] below.
For all your answers, please write down your calculations and explain your reasoning. Draw
sketches whenever helpful, and please use your units correctly. If you are using any online
resources (such as unit conversion tools) give their URL and webpage title in your answers.
[a] For Example 10.13 (Calculating the Torque in a Kick) in Section 10.5, replace “2000-N force"
with "X-N force" where X is the number you computed on the first page of this study guide.
Then solve part (a) of the example. There is no need to solve part (b).
X = 883
%3D

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