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Sidana [21]
3 years ago
11

A turntable must spin at 33.4 rpm (3.50 rad/s) to play an old-fashioned vinyl record. how much torque must the motor deliver if

the turntable is to reach its final angular speed in 1.0 revolutions, starting from rest? the turntable is a uniform disk of diameter 30.5 cm and mass 0.21 kg.
Physics
1 answer:
Westkost [7]3 years ago
4 0
In order to compute the torque required, we may apply Newton's second law for circular motion:
Torque = moment of inertia * angular acceleration

For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds

α = (3.5 - 0) / 1.8
α = 1.94 rad/s²

The moment of inertia of a thin disc is given by:
I = MR²/2
I = (0.21*0.1525²)/2
I = 0.002

τ = 1.94 * 0.002
τ = 0.004

The torque is 0.004
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3 years ago
A fast moving vehicle travelling at a speed of 25.4 m/s comes up behind another vehicle which is
ale4655 [162]

Answer:

     a = 6.1 m / s²

Explanation:

For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle

Let's find the relative initial velocity of the two vehicles

          v₀ = v₀₂ - v₀₁

          v₀ = 25.4 - 13.6

          v₀ = 11.8 m / s

 

the fastest vehicle

          x = v₀ t + ½ a t²

The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is

          x = 11.4 m

         

let's use the expression

           v² = v₀² - 2 a x

how the vehicle stops v = 0

          a = v₀² / 2x

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this velocity is directed to the left

8 0
3 years ago
A 1000 Kg car approaches an intersection traveling north at 30 m/s . A 1250 Kg car approaches the same intersection traveling ea
Feliz [49]

Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, m_1=1000\ kg

Initial velocity of first car is, u_1=30\vec{j}\ m/s

Mass of second car is, m_2=1250\ kg

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, u_{2y}=0.

The magnitude of final velocity is given as:

|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

7 0
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gladu [14]

Answer:

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