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Talja [164]
3 years ago
6

At a certain instant a rigid wheel is spinning about its center of mass with angular velocity of magnitude ω and angular acceler

ation of magnitude α. Consider a point a distance r from the axis of rotation. What would be the magnitude of the acceleration of this point on the wheel?
Physics
1 answer:
geniusboy [140]3 years ago
7 0

Answer:

a = r√(w⁴ + (alpha)²)

Explanation:

let w be the magnitude of angular velocity

r = radius

v = velocity = wr

angular acceleration, ā = v²/r = (wr)²/r = w²r

Also,

â = dv/dt = rdw/dt

given : alpha = dw/dt

hence, â = (alpha)r

the resultant acceleration, a, by Pythagoras is given as

a = √( â² + ā²) = √(r²w⁴ + r²(alpha)²)

simplifying,

a = √r²(w⁴ + (alpha)²)

a = r√(w⁴ + (alpha)²)

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
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Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
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Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

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