The flywheel of a steam engine runs with a constant angular speed of 169 rev/min. When steam is shut off, the friction of the bearings and the air bring the wheel to rest in 2.9 h. What is the magnitude of the constant angular acceleration of the wheel?

How many rotations does the wheel make before coming to rest?

What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 35 cm from the axis of rotation?

What is the magnitude of the net linear acceleration of the particle in the above question when the flywheel is turning at 84 rev/min?

Explanation from Alloprof

This Explanation was submitted by a member of the Alloprof team.

Hi HibouHabile8386 😊

Thank you for your question!

The formula for angular expression is the following:

$$ a = \frac{v^2}{r} = \frac{∆v}{t} $$

Legend :

• a : angular acceleration

• v : angular speed

• r : radius

• ∆v : angular speed variation

• t : time (minutes)

Therefore, we can find the answer to the problem by dividing the change in speed over the deceleration period:

$$ a = \frac{-169\:rev/min}{2.9\:h•60\:min} ≈ -0.971\:rev/min $$

To find the number of rotations the wheel makes, you can use this equation :

$$ n = v•t + 1/2•a•t^2 $$

$$ n = 169•(2.9•60) + 1/2•(-0.971)•(2.9•60)^2 $$

Unfortunately, your other questions exceed the difficulty of what is taught in Quebec high schools. Despite wanting to help, we are thus unable to answer them.

Don't hesitate to ask for more help!