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Naddika [18.5K]
3 years ago
13

A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicul

ar to the rod. A horizontal spring, assumed massless andwith force constant k,is attached to the lower end of the rod, with the other end of thespring attached to a rigid support.
Find the torque tau due to the spring. Assume that theta is small enough that the spring remains effectivelyhorizontal and you can approximate \sin{(\theta)}\approx \theta (and \cos(\theta)\approx 1 ).
Express the torque as a function oftheta and other parameters of the problem.
What is the angular frequency omega of oscillations of the rod?
Express the angular frequency interms of parameters given in the introduction.
Physics
2 answers:
lys-0071 [83]3 years ago
8 0

Answer:

Explanation:

Moment of inertia of the metal rod pivoted in the middle

= M l² / 12

If the spring is compressed by small distance x twisting the rod by angle θ

restoring force by spring

= k x

moment of torque  about axis

= k x l /2

= k θ( l /2 )²     ( x / .5 l = θ )

=

moment of torque = moment of inertia of  rod  x angular acceleration

k θ( l /2 )²   = M l² / 12 d²θ/dt²

d²θ/dt² = 3 k/M  θ

acceleration =  ω² θ

ω² = 3 k/M

ω = √ 3 k / M

andrezito [222]3 years ago
3 0

Answer:

τ = FR = (−kRθ)R = −kR²θ

ω=√(3k/M)

Explanation:

the rod rotates by a small angle θ.  

The spring stretches by amount x = Rθ, and to exert a force

F = −kx = −kRθ on the rod.

Torque equal to

τ = FR = (−kRθ)R = −kR²θ. .........................1

Now, we know that τ = Iα (torque = moment of inertia times angular acceleration);

equsating 1 with the value of torque

Iα = −kR²θ .....................ii

moment of inertia of a rod is given by

I = ML²/12 = M(2R)²/12 = MR²/3 ...................................2

Substituting 2 into the equ 2

(MR²/3)α = −kR²θ

or:

Mα/3 = −kθ

Now, since α is the 2nd derivative of θ (with respect to time), this is just a differential equation:

(M/3)θ′′ = −kθ

from simple harmonic equation

we know that

T=2π√(m/k)

, "mx′′ = −kx"; except that "M/3" takes the place of "m".  That means, where the period of the standard SHM equation comes out to T=2π√(m/k),

we substitute M/3 for m

T = 2π√(M/(3k))..............3

ω=2π/T........4

T=2π/ω

putting angular frequency into 3

ω=√(3k/M)

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An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v
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Given:
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Read 2 more answers
Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
s2008m [1.1K]

Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

after the collision we have that

p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

so

p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

Putting in one side of the equation each speed we get

\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the  speed of the 9m particle:

v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

the magnitude of the final speed of the particle 9m is

v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

The tangent that the speed of the particle 9m makes with the -y axis is

tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}

As a vector the speed of the particle 9m is:

\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}

As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

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3 years ago
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andreyandreev [35.5K]

The apparent weight of a 1.1 g drop of water is 4.24084 N.

<h3>What is Apparent Weight?</h3>
  • According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
  • By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.

Solution:

N = Speed of rotation = 1250 rpm

D = Diameter = 45 cm

r = Radius = 22.5 cm

M = Mass of drop = 1.1 g

Angular speed of the water = \omega  = \frac{2\pi N}{60}

\omega  = \frac{2\pi \times 1250}{60}

\omega  = 130.89 rad/s

Apparent weight is given by

W _a = M\omega^{2}R

W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225

W_a = 4.24084 N

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Question:

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?

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