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2 years ago

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A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N t

o the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. What is the wheels angular acceleration? What force does the axle apply to the wheel? What torque does the axle apply to the wheel?

1 answer:

Grace [21]2 years ago

3 0

Answer: $25.97 rad/s^2$

Explanation:

Given

radius of wheel $r=0.28 m$

mass of wheel $m=8.80 kg$

Force $F=32 N$

Moment of Inertia of solid wheel $I=\frac{mr^2}{2}$

$I=\frac{8.8\times 0.28^2}{2}$

$I=0.344 kg-m^2$

Torque is given by

$\tau =F\times r=I\times \alpha$

$32\times 0.28=0.344\times \alpha$

$\alpha =25.97 rad/s^2$

Force on the axle is 32 N since there is no linear acceleration of the system

using Third law F=32 N

Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle

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Answer:

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Explanation:

The given data :-

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$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

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Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

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Surface charge density ( σ₂ ) for smaller sphere.

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Answer:

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