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2 years ago

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A string is wrapped around a disk of mass 1.7 kg and radius 0.11 m. Starting from rest, you pull the string with a constant forc

e 7 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance 0.15 m, your hand has moved a distance of 0.22 m.
a. At this instant, what is the speed of the center of mass of the disk?
b. At this instant, how much rotational kinetic energy does the disk have relative to its center of mass?

2 answers:

V125BC [204]2 years ago

5 0

Answer:

Explanation:

a) work done in moving the disk by 0.15 m = its kinetic energy = 0.5 mv²

Fd = 0.5 mv²

2 fd = mv²

√( $\frac{2fd}{m}$ ) = v

v = 1.11 m/s

b) rotational kinetic energy = F ( 0.22 m - 0.15 m) = 0.49 J

Evgen [1.6K]2 years ago

5 0

Answer:

(a) Vcm = 1.43m/s

(b) KEcm = 2.59J

Explanation:

Given

m = mass of the solid disk = 1.7kg

Radius R = 0.11m

F = 7N

Let the distance moved by the the center of mass be x = 0.15m

And the distance moved by in unwinding the rope be d = 0.22m

The total workdone on the disk is causing it to rotate and also move its center of mass is F×(d+x)

W = 7×(0.15+0.22)

= 7× 0.37 = 2.59J

By work-energy theorem,

W = ΔKE = 1/2mVcm² + 1/2×I×ω²....(1)

I = moment of inertia = 1/2MR² and

ω = angular speed = Vcm/R

So substituting these expressions into the equation above we have

W = 1/2MVcm² + 1/2×1/2MR²×(Vcm/R)²

W = 1/2MVcm² + 1/4MVcm²

W = 3/4MVcm²

Vcm² = 4/3×W/M

Vcm = √(4/3×W/M)

Vcm = √(4/3×2.59/1.7)

Vcm = 1.43m/s

KE = workdone = 2.59J.

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A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =

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A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

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1nC= 10⁻⁹C

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Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

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$d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }$

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(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

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F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N

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x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

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Fn direction (α)

$\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )$

$\alpha =tan^{-1}( \frac{-3.67 }{5.71} )$

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

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