Question

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switched off, it coasts to rest in 32 seconds. Determine the number of revolutions turned during both the startup and shutdown periods. Also determine the number of revolutions turned during the first half of each period. Assume uniform angular acceleration in both cases.

Answer 1

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

• Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       $\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$  

• Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       $\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

• Since the machine starts from rest, ω₀ = 0.
• We know the value of ωf₁ (the operating speed) in rev/min.
• Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       $3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

• Replacing by the givens in (2):

       $57.5 rev/sec = 0 + \alpha * 6 s (4)$

• Solving for α:

       $\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

• Replacing (5) and Δt in (1), we get:

       $\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

• in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       $\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

• In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       $\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

• First of all, we need to find the value of the angular acceleration during the second period.
• We can use again (2) replacing by the givens:
• ωf =0 (the machine finally comes to an stop)
• ω₀ = ωf₁ = 57.5 rev/sec
• Δt = 32 s

       $0 = 57.5 rev/sec + \alpha * 32 s (9)$

• Solving for α in (9), we get:

       $\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

• Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        $\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

• In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
• $\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$