Question

I start pushing a merry-go-round of radius of 1.9 meters with a
tangential force of 29.8 N It has a moment of inertia of 187.7 kg
m^{2}. What is its rotational speed in rad/s after 3.85
seconds assuming it starts at rest?

Answer #1

Radius of merry go round r = 1.9m.

Moment of inertia I = 187.7kg•m².

Tangential force F = 29.8N.

Let angular acceleration is **α .**

Now find torque about the center and balance them.

So formula for torque

T = F×r×sin∅

= 29.8×1.9×sin90° (as force is tangentially so angle between force and radius is 90°)

T = 56.62 Nm

Now formula of torque in the term of moment of inertia and angular acceleration.

T = I×**α**

**s**o I×α = 56.62

α = 56.62/187.7

So angular acceleration ×α = 0.3rad/s²

Now formula for angular speed v

v = u + t×α { where u is initial speed in our 0 and t is time = 3.85s}

v = 3.85×0.3

V = 1.16rad/s

So rotational speed after 3.85s is 1.16rad/s.

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