4. A 180 kg sumo wrestler on roller skates, rolling at 5 m/s to the left, collides elastically with

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

choli [55]

Answer:

They can be seen from a distance of 4.372 kilometers.

Explanation:

Using the Reyligh creterion for diffraction through a circular aperture we have

$\frac{x}{D}=\frac{1.22\lambda }{d}$

where symbol's have their usual meaning

thus applying values we get

$D=\frac{dx}{1.22\lambda }$

$\therefore D=\frac{0.633\times 4.61\times 10^{-3}}{1.22\times 547\times 10^{-9}}\\\\D=4372.77m\\=4.372km$

5 0

3 years ago

You dribble a basketball while walking on a basketball court. List and describe at least 3

Airida [17]

Answer:

1. You push on the ball and the ball pushes on your hand .

2. The ball hits the ground and the ground pushes back on the ball .

3. You walk on the ground with your feet and the ground pushes back on you.

Explanation:

4 0

3 years ago

Read 2 more answers

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

3 years ago

Light of wavelength 480 nm illuminates a pair of slits separated by 0.27 mm. If a screen is place 1.7 m from the slits, determin

Dahasolnce [82]

Answer:

Δy = 6.05 mm

Explanation:

The double slit phenomenon is described by the expression

d sin θ = m λ constructive interference

d sin θ = (m + ½) λ destructive interference

m = 0,±1, ±2, ...

As they tell us that they measure the dark stripes, we are in a case of destructive interference, let's use trigonometry to find the sins tea

tan θ = y / x

y = x tan θ

In the interference experiments the measured angle is very small so we can approximate the tangent

tan θ = sin θ / cos θ

cos θ = 1

tan θ = sin θ

y = x sin θ

We substitute in the destructive interference equation

d (y / x) = (m + ½) λ

y = (m + ½) λ x / d

The first dark strip occurs for m = 0 and the third dark strip for m = 2. Let's find the distance for these and subtract it

m = 0

y₀ = (0+ ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

y₀ = 1.511 10⁻³ m

m = 2

y₂ = (2 + ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

y₂ = 7.556 10⁻³ m

The separation between these strips is Δy

Δy = y₂-y₀

Δy = (7.556 - 1.511) 10⁻³

Δy = 6.045 10⁻³ m

Δy = 6.05 mm

5 0

3 years ago