Ask question

Ask question

All categories

1 month ago

14

When switched on, the grinding machine accelerates from rest to its operating speed of 3550 rev/min in 10 seconds. When switched

off, it coasts to rest in 31 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.

1 answer:

serg [3.5K]1 month ago

5 0

Response:

At startup, the total revolutions

$N_1 = 296 turn$

After the first half of the duration, total revolutions

$N_3 = \frac{29.6 + 0}{2}(5) = 74 turns$

Total revolutions until it halts

$N_2 = \frac{59.17 + 0}{2}(31) = 917 turn$

Following the first half of the duration, the total revolutions are

$N_4 = 688 turns$

Clarification:

At the beginning, the machine is stationary and then accelerates to a speed of 3550 rev/min

At this point we calculate

$f = \frac{3550}{60} = 59.17 rev/s$

Since it took 10 seconds to achieve the speed

the angular acceleration can be expressed as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{59.17- 0}{10} = 5.917 rev/s^2$

It comes to a stop after 31 seconds, leading to the angular deceleration being defined as

$\alpha_2 = -\frac{59.17}{31} = -1.91 rev/s^2$

Initially, the revolutions taken to reach the specified speed

$N_1 = \frac{59.17 + 0}{2}(10) = 296 turn$

the number of revolutions during the stopping phase is

$N_2 = \frac{59.17 + 0}{2}(31) = 917 turn$

Now, considering the increase in speed after t = 5 seconds is

$\omega_1 = (5.917)(5) = 29.6 rev/s$

$N_3 = \frac{29.6 + 0}{2}(5) = 74 turns$

During the stopping phase, the speed after half of the duration is calculated as

$\omega_2 = 59.17 - (1.91)(15.5) = 29.56 rev/s$

Lastly, the total revolutions are calculated as

$N_4 = \frac{59.17 + 29.56}{2}(15.5)$

$N_4 = 688 turns$

You might be interested in

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

Softa [3030]

A. A car moving at a constant speed on a flat, straight road. B. A vehicle traveling at a steady speed on a 10-degree incline. An object operates within an inertial reference frame if there is no net force acting upon it. According to Newton's second law, this implies that the object's acceleration also equals zero. Assessing the scenarios yields: A. A car moving at a constant speed on a flat road qualifies as an inertial reference frame, since its velocity and direction remain unchanged; thus, acceleration is zero. B. A car moving steadily up a 10-degree incline still constitutes an inertial reference frame, for similar reasons. C. A car accelerating after departing a stop sign does not represent an inertial frame due to its change in speed. D. A car driving at a steady speed around a curve cannot be considered an inertial reference frame since its direction is changing, resulting in a change in velocity and thus acceleration. Therefore, options A and B are correct.

8 0

1 month ago

Read 2 more answers

A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part

ValentinkaMS [3465]

answer:

$Let frictional forces due to both hands and feets be "Ff" each(since its given that they all are equal), acting in upward direction( in opposite direction of supposed motion).\\
Then since there is no motion of gymnast thus net frictional force due to both hands and feets will be exactly balanced by the weight of the gymnast,\\ i.e\\
4f_{f}=weight =mg\\
f_{f}=\frac{55x9.8}{4}\\
=134.75N$

5 0

2 months ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

3 months ago