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P = d W d t .

If we have a constant net torque, [link] becomes W = τ θ and the power is

P = d W d t = d d t ( τ θ ) = τ d θ d t

or

P = τ ω .

Torque on a boat propeller

A boat engine operating at 9.0 × 10 4 W is running at 300 rev/min. What is the torque on the propeller shaft?

Strategy

We are given the rotation rate in rev/min and the power consumption, so we can easily calculate the torque.

Solution

300.0 rev/min = 31.4 rad/s;
τ = P ω = 9.0 × 10 4 N · m / s 31.4 rad / s = 2864.8 N · m .

Significance

It is important to note the radian is a dimensionless unit because its definition is the ratio of two lengths. It therefore does not appear in the solution.

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Check Your Understanding A constant torque of 500 kN · m is applied to a wind turbine to keep it rotating at 6 rad/s. What is the power required to keep the turbine rotating?

3 MW

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Rotational and translational relationships summarized

The rotational quantities and their linear analog are summarized in three tables. [link] summarizes the rotational variables for circular motion about a fixed axis with their linear analogs and the connecting equation, except for the centripetal acceleration, which stands by itself. [link] summarizes the rotational and translational kinematic equations. [link] summarizes the rotational dynamics equations with their linear analogs.

Rotational and translational variables: summary
Rotational Translational Relationship
θ x θ = s r
ω v t ω = v t r
α a t α = a t r
a c a c = v t 2 r
Rotational and translational kinematic equations: summary
Rotational Translational
θ f = θ 0 + ω t x = x 0 + v t
ω f = ω 0 + α t v f = v 0 + a t
θ f = θ 0 + ω 0 t + 1 2 α t 2 x f = x 0 + v 0 t + 1 2 a t 2
ω f 2 = ω 2 0 + 2 α ( Δ θ ) v f 2 = v 2 0 + 2 a ( Δ x )
Rotational and translational equations: dynamics
Rotational Translational
I = i m i r i 2 m
K = 1 2 I ω 2 K = 1 2 m v 2
i τ i = I α i F i = m a
W A B = θ A θ B ( i τ i ) d θ W = F · d s
P = τ ω P = F · v

Summary

  • The incremental work dW in rotating a rigid body about a fixed axis is the sum of the torques about the axis times the incremental angle d θ .
  • The total work done to rotate a rigid body through an angle θ about a fixed axis is the sum of the torques integrated over the angular displacement. If the torque is a constant as a function of θ , then W A B = τ ( θ B θ A ) .
  • The work-energy theorem relates the rotational work done to the change in rotational kinetic energy: W A B = K B K A where K = 1 2 I ω 2 .
  • The power delivered to a system that is rotating about a fixed axis is the torque times the angular velocity, P = τ ω .

Key equations

Angular position θ = s r
Angular velocity ω = lim Δ t 0 Δ θ Δ t = d θ d t
Tangential speed v t = r ω
Angular acceleration α = lim Δ t 0 Δ ω Δ t = d ω d t = d 2 θ d t 2
Tangential acceleration a t = r α
Average angular velocity ω = ω 0 + ω f 2
Angular displacement θ f = θ 0 + ω t
Angular velocity from constant angular acceleration ω f = ω 0 + α t
Angular velocity from displacement and
constant angular acceleration
θ f = θ 0 + ω 0 t + 1 2 α t 2
Change in angular velocity ω f 2 = ω 0 2 + 2 α ( Δ θ )
Total acceleration a = a c + a t
Rotational kinetic energy K = 1 2 ( j m j r j 2 ) ω 2
Moment of inertia I = j m j r j 2
Rotational kinetic energy in terms of the
moment of inertia of a rigid body
K = 1 2 I ω 2
Moment of inertia of a continuous object I = r 2 d m
Parallel-axis theorem I parallel-axis = I initial + m d 2
Moment of inertia of a compound object I total = i I i
Torque vector τ = r × F
Magnitude of torque | τ | = r F
Total torque τ net = i | τ i |
Newton’s second law for rotation i τ i = I α
Incremental work done by a torque d W = ( i τ i ) d θ
Work-energy theorem W A B = K B K A
Rotational work done by net force W A B = θ A θ B ( i τ i ) d θ
Rotational power P = τ ω
Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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