Question

Consider a motor that exerts a constant torque of 25.0N \cdot m to a horizontal platform whose moment of inertia is 50.0kg \cdot m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.

Part A ) How much work W does the motor do on the platform during this process?
Enter your answer in joules to four significant figures.
W =
1885
\rm J
Part B ) What is the rotational kinetic energy of the platform K_rot,f at the end of the process described above?
Enter your answer in joules to four significant figures.
K_rot,f =
1885
\rm J
Part C ) What is the angular velocity omega_f of the platform at the end of this process?
Enter your answer in radians per second to three significant figures.
omega_f =
8.68
{\rm rad / s}
Part D ) How long \Delta t does it take for the motor to do the work done on the platform calculated in Part A?
Enter your answer in seconds to three significant figures.
\Delta t =
17.4
\rm s
Part E ) What is the average power P_avg delivered by the motor in the situation above?
Enter your answer in watts to three significant figures.
P_avg = 109 \rm W
Part F ) Note that the instantaneous power P delivered by the motor is directly proportional to omega, so P increases as the platform spins faster and faster. How does the instantaneous power P_f being delivered by the motor at the time t_{\rm f} compare to the average power P_avg calculated in Part E?
Note that the instantaneous power delivered by the motor is directly proportional to , so increases as the platform spins faster and faster. How does the instantaneous power being delivered by the motor at the time compare to the average power calculated in Part E?
P = P_{\rm avg}
P = 2 * P_{\rm avg}
P = P_{\rm avg} / 2
none of the above

Answer 1

Answer:

A)  W = 1885 J , B) $K_{f}$ = 1885 J , C)  w = 8.68 rad / s , D)  t = 8,687 s , E)  P = 109 W  F) P = 2 $P_{rms}$

Explanation:

Part A    The work in the rotational movement is

       W = τ θ

Let's look at the rotated angle

      θ = 12.0 rot (2pi rad / 1rot) = 75.398 rad

     

     W = 25.0 75.40

     W = 1885 J

Part B   Let's use the relationship between work and kinetic energy

      W = ΔK = Kf - Ko

As the body leaves the rest w₀ = 0     ⇒ K₀ = 0

      W = $K_{f}$ -0

      $K_{f}$ = 1885 J

Part C     The formula for kinetic energy is

      K = ½ I w²

     w² = 2k / I

     w = √ (2 1885/50)

     w = 8.68 rad / s

Part D     The power in the rotational movement

     P = τ w

     P = 25 8.68

     P = 217 W

     

     P = W / t

     t = W / P

     t = 1885/217

     t = 8,687 s

Part E   At average power is

     P = τ ($w_{f}$ -w₀)/ 2

We look for angular velocity with kinematics

    $w_{f}[/tex = w₀ + α t

     

     τ = I α

      α = τ / I

      α = 25/50

      α = 0.5 rad / s²

calculate

      P = 25 (0.5 8.687)

      P = 108.6 W

      P = 109 W

Part F    

The average power is

      [tex]P_{rms}$ = τ ($w_{f}$ -w₀) /

The instant power is

      P = τ w

The difference is that in one case the angular velocity is instantaneous and between averages

P / $P_{rms}$ = τ w / (τ ($w_{f}$-w₀) / 2)

P / $P_{rms}$= 2 w / Δw

For this case w₀ = o

p / $P_{rms}$ = 2