Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is 
Explanation:
From the question we are told that
The mass of the steel ball is 
The length of arm is 
The mass of the arm is 
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

=>
=>
Generally the magnitude of torque about the athlete shoulder is mathematically represented as

=> 
=> 
(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
<h3>
Free body diagram</h3>
The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;
/ W2
Ф → Ff
↓W1
where;
- Ff is the frictional force resisting the down motion of the box
- W1 is the perpendicular component of the box weight = Wcos(33)
- W2 is the parallel component of the box weight = Wsin(33)
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
Learn more about free body diagram of inclined objects here: brainly.com/question/4176810
A group of individuals living in a particular geographic area is termed population.
The new oscillation frequency of the pendulum clock is 1.14 rad/s.
The given parameters;
- <em>Mass of the pendulum, = M </em>
- <em>Length of the pendulum, = L</em>
- <em>Initial angular speed, </em>
<em> = 1 rad/s</em>
The moment of inertia of the rod about the end is given as;

The moment of inertia of the rod between the middle and the end is calculated as;
![I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}](https://tex.z-dn.net/?f=I_f%20%3D%20%5Cint%5Climits%5EL_%7BL%2F2%7D%20%7Br%5E2%5Cfrac%7BM%7D%7BL%7D%20%7D%20%5C%2C%20dr%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5Br%5E3%5D%5EL_%7BL%2F2%7D%20%3D%20%20%5Cfrac%7BM%7D%7B3L%7D%20%5BL%5E3%20-%20%5Cfrac%7BL%5E3%7D%7B8%7D%20%5D%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5B%5Cfrac%7B7L%5E3%7D%7B8%7D%20%5D%3D%20%5Cfrac%7B7ML%5E2%7D%7B24%7D)
Apply the principle of conservation of angular momentum as shown below;

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.
Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129
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