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BaLLatris [955]
3 years ago
7

A particular motor can provide a maximum torque of 110 n • m. assuming that all of this torque is used to accelerate a solid, un

iform, cylindrical flywheel of mass 10.0 kg and radius 3.00 m, how long will it take for the flywheel to accelerate from rest to 8.13 rad/s?
Physics
1 answer:
hjlf3 years ago
5 0

Using the rotational equivalent of force:

Which is T = I*Alpha

Where: T is torque, I is the moment of inertia and Alpha is the angular acceleration. 

This is for the flywheel: J = 1/2mr^2 = 5*3^2 = 45 kgm^2 


From the equation: 


T = J*dω/dt 


we get: 


Δt = J*Δω/T = 45*8.13/110.0 = 3.326 sec

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v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

7 0
3 years ago
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Answer:

-2.67 m/s²

Explanation:

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