Suppose you exert a force of 182 N tangential to a 0.33 m radius, 79 kg grindstone (a solid disk). What is the angular acceleration (in SI units) if there is an opposing frictional force of 24 N exerted 0.17 m from the axis?

Enter only the number of your answer with 3 or more significant figures.

Tangential force exerted is 182 Newton at a distance of 0.33 m. The mass of the disc is 79 kg the opposing frictional force is of 24 Newton which is exerted distance of 0.7 m from the axis.

We know that the angular acceleration is given by,

A = total net torque/moment of inertia

A is angular acceleration.

Torque = tangential force × distance

Torque by force = 182 x 0.33

Torque by force = 60.06 N-m

Torque by friction = 24 x 0.17

Torque by friction = 4.08 N-m

The net torque = (60.06-4.08)N-m

Moment of inertia of disc = 79(0.33)²/2

Moment of inertia of disc = 4.301 Kg-m²

Now putting all the values,

A = 55.98/4.301

A = 13.015 rad/s²

So, the angular acceleration is 13.015rad/s².

To know more about angular acceleration, visit,

https://brainly.com/question/21278452

#SPJ4

Suppose you exert a force of 182 N tangential to a 0.33 m radius, 79 kg grindstone (a solid disk). What is the angular acceleration (in SI units) if there is an opposing frictional force of 24 N exerted 0.17 m from the axis? Enter only the number of your answer with 3 or more significant figures.

Answer :

Other Questions

Suppose you exert a force of 182 N tangential to a 0.33 m radius, 79 kg grindstone (a solid disk). What is the angular acceleration (in SI units) if there is an opposing frictional force of 24 N exerted 0.17 m from the axis?

Enter only the number of your answer with 3 or more significant figures.