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Welcome to our Physics lesson on **Power in Rotational Motion**, this is the eighth lesson of our suite of physics lessons covering the topic of **Dynamics of Rotational Motion**, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

The last quantity in which the analogy between translational and rotational quantities is valid, is **power**. Thus, given that power in translational motion is

Power = *Work**/**time*

=*F**⃗* × ∆x*⃗**/**t*

= F*⃗* × v*⃗*

=

= F

we obtain for power in rotational motion

P_{rot} = *W*_{rot}*/**t*

=*τ × φ**/**t*

= τ × ω

=

= τ × ω

The end of a 60 cm and 300 g rod start rotating from rest rotates in the horizontal plane through a small electric motor around a vertical axis as shown in the figure.

10 s after its start of rotation, the rod gains an angular velocity of 5 rad/s.

- Work done by the electric motor to rotate the rod
- Angular momentum at the end of 10 s
- Rotational kinetic energy of the rod at the end of 10 s
- Average power delivered by the motor supposing that all energy produced by it, is used to do work.

**a.** We must calculate torque and angle of rotation to find rotational work, as he Newton's Second Law of rotational motion is

τ = I × α

Moment of inertia I of a bar rotating around its end is

I = *1**/**3* m × L^{2}

where m = 300 g = 0.3 kg and L = 60 cm = 0.6 m. Thus,

I = *1**/**3* × 0.3 kg × (0.6 m)^{2}

= 0.036 kg × m^{2}

= 0.036 kg × m

Angular acceleration α is calculated by

α = *ω - ω*_{0}*/**t*

where ω = 5 rad/s, ω_{0} = 0 and t = 10 s. Thus,

α = *5 **rad**/**s* - 0 *rad**/**s**/**10 s*

= 0.5*rad**/**s*^{2}

= 0.5 s^{-2}

= 0.5

= 0.5 s

Hence,

τ = I × α

= 0.036 kg × m^{2} × 0.5 s^{-2}

= 0.018*kg × m*^{2}*/**s*^{2}

= 0.018 N × m

= 0.036 kg × m

= 0.018

= 0.018 N × m

(Remember that 1 N = 1 kg × m/s^{2})

The angle φ wiped by the rod during its rotation is

φ = ω_{0} × t + *α × t*^{2}*/**2*

= 0 × 10 +*0.5 × 10*^{2}*/**2*

= 25 rad

= 0 × 10 +

= 25 rad

Hence, the rotational work done by the engine is

W_{rot} = τ × φ

= 0.018 N × m × 25 rad

= 0.45 J

= 0.018 N × m × 25 rad

= 0.45 J

**b.** Angular momentum at the end of 10 s is

L = I × ω

Substituting the values we found earlier, we obtain

L = 0.036 kg × m^{2} × 5 *rad**/**s*

= 0.18*kg × m*^{2}*/**s*

= 0.18

**c.** Rotational kinetic energy of the rod at the end of 10 s is

KE_{rot} = *I × ω*^{2}*/**2*

=*0.036 kg × m*^{2} × (5 *rad**/**s*)^{2}*/**2*

= 0.45 J

=

= 0.45 J

As you see, this value is equal to the work done by the motor to make the rod rotate.

**d.** Average power delivered by the motor supposing that all energy produced by it, is used to do work, is:

P_{rot} = *W*_{rot}*/**t*

=*0.45 J**/**10 s*

= 0.045 W

=

= 0.045 W

The following table includes everything discussed above regarding the relationship between dynamics translational and rotational quantities and the relevant formulae:

You have reached the end of Physics lesson **7.2.8 Power in Rotational Motion**. There are 8 lessons in this physics tutorial covering **Dynamics of Rotational Motion**, you can access all the lessons from this tutorial below.

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