Question

A gymnast performing on the high bar can be modeled as a rod rotating around its end.

Answer 1

The magnitude of the torque is $1466 N\cdot m$

Explanation:

The torque exerted by a force acting perpendicular to the arm is given by:

$\tau = F d$

where

F is the magnitude of the force

d is the distance between the point of application of the force and the pivot

In this problem, the force acting on the gymnast is his weight, so the magnitude is given by

$F=mg = (68 kg)(9.8 m/s^2)=666.4 N$

where

m = 68 kg is the mass of the gymnast

$g=9.8 m/s^2$ is the acceleration of gravity

Now we can use the first equation to find the torque, where

d = 2.2 m

And substituting,

$\tau = (666.4)(2.2)=1466 N\cdot m$

Learn more about torque and levers:

brainly.com/question/5352966

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