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Gnesinka [82]
3 years ago
14

A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the cen

ter of the sphere is to be 0.10 gb) For this angle, what would be the acceleration of a frictionless block sliding down the incline
Physics
2 answers:
Liono4ka [1.6K]3 years ago
5 0

Answer:

Part a)

\theta = 8.05 degree

Part b)

a = 1.37 m/s^2

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

mg sin\theta - F_f = ma

also we have torque equation on it

F_f R = I\alpha

for pure rolling

a = R \alpha

F_f = \frac{Ia}{R^2}

now we have

mg sin\theta = ma + \frac{Ia}{R^2}

now we have

mg sin\theta = (m + \frac{2}{5}m)a

a = \frac{5}{7}g sin\theta

now given that

a = 0.10 g

so we have

0.10 g = \frac{5}{7} g sin\theta

sin\theta = 0.14

\theta = 8.05 degree

Part b)

If the inclined plane is frictionless then the acceleration is given as

a = g sin\theta

a = 9.8(0.14)

a = 1.37 m/s^2

velikii [3]3 years ago
5 0

Answer:

Explanation:

Given

Moment of inertia of sphere is I=\frac{2Mr^2}{5}

Here Friction will Provide torque

thus f_r\times r=I\times \alpha

f_r=\frac{I}{r^2}

Also friction will oppose weight sin component

mg\sin \theta -f_r=ma

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

a=0.1 g=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

\sin \theta =0.14

\theta =8.407^{\circ}

(b)Considering \mu the coefficient of friction

acceleration of block sliding down the incline

a=g\sin \theta -\mu g\cos \theta

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Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}    .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}    .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

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A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. However, he is unable to lift the suit
kvasek [131]

Answer:

b)

Explanation:

Normal force, is always directed upward the surface over which is placed the object, and can adopt any value, as required to meet Newton's 2nd Law.

In this case, as the external force on the suitcase pulls upward, in order  to counteract the influence of gravity, normal force is less than the weight of the suitcase, as follows:

F + Fn = m*g

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So, the normal force is equal to the magnitude of the weight of the suitcase (m*g) minus the magnitude of the force of the pull (F) which is the same expressed by the statement b.

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Which would have the longer orbital period: a moon 1 million km from the center of Jupiter, or a moon 1 million km from the cent
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Answer:

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Explanation:

The formula for the orbital period of the moon is given by

T = 2\pi \sqrt{\frac{r}{g}}

As the time period is inversely proportional to the square root of the acceleration due to gravity of the planet.

As the value of acceleration due to gravity on Jupiter is more than the earth, so the period of moon around the earth is large as compared to the period of the moon around the Jupiter when the distance is same.

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Starting from rest, a 68.0 kg woman jumps down to the floor from a height of 0.790 m, and immediately jumps back up into the air
Aleonysh [2.5K]

Answer:

a) I = 0 N s,  b)  v = -3.935 m / s, c) vf = 3.935 m / s,   d)   y = 0.790 m

Explanation:

a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship

          I = ∫F dt

The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction

        I = ∫ (9200 t - 11500 t2) dt

        I = 9200 t² / 2 - 11500 t³ / 3

We evaluate between the lower limit t = 0 and upper limit t = 0.800 s

       I = 9200 (0.8² -0) - 11500 (0.8³ -0)

       I = 5888 -5888

       I = 0 N s

Directed from the floor to the woman

b) For this part we use kinematics

      v² = v₀² - 2g y

     v = √ (0 - 2 9.8 (-0.79))

     v = 3.935 m / s

The speed direction is down

c) for this we use the relationship between momentum and the amount of movement

      I = ΔP

      I = m vf - m v₀

     vf = (I + m v₀) / m

This is the impulse of women on the floor    

      vf = ( 0 + 68 (3.935)) / 68

      vf = 3.935 m / s

d) let's use kinematics

      v₂ = v₀² - 2gy

      0 = v₀² - 2gy

      y = v₀² / 2g

      y = 3.935²/2 9.8

      y = 0.790 m

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