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Svetach [21]
3 years ago
6

26. A solid wheel accelerates at 3.25 rad/s2 when a

Physics
1 answer:
Yanka [14]3 years ago
5 0

Answer:

9.0 N

Explanation:

The location of the mass of the wheel on the wheel = Evenly distributed

The acceleration of the solid wheel, α = 3.25 rad/s²

The applied force on the wheel = 4.5 N

The location mass of the replacement wheel = All on (around) the rim

The moment of inertia of the new wheel, I = m·r² (From an online source)

We have;

The moment of inertia for a solid wheel = 1/2·m·r²

The torque, τ = Moment of inertia, I × Acceleration, α

For the solid wheel, we have;

τ  = 1/2·m·r² × 3.25 rad/s²

τ = r × F = r × m × a

For the replacement wheel, we have;

τ  = m·r² × 3.25 rad/s² = 2 × 1/2·m·r² × 3.25 rad/s²

∴ τ = 2 × r × F

Given that the radius remains the same, the force applied on the replacement wheel needs to be doubled

The force that should be exerted on the strap to give the same angular velocity, F' = 2 × F

The required force, F' = 2 × 4.5 N = 9.0 N.

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Answer:

Explanation:

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The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

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The normal force exerted at B is 0.69N

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Ft=mv²/r

Then,

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Fa=mv²/r

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3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

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Vb=3.22m/s

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W=∆K.E+P.E

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The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

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