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Paul [167]
2 years ago
8

A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can r

otate with negligible friction about a stationary horizontal axis. The drum is not a uniform cylinder and has unknown moment of inertia. When you release the bucket from rest, you find that it has a downward acceleration of magnitude (a). What is the tension in the cable between the drum and the bucket? (b) What is the moment of inertia of the drum about its rotation axis?
c) In the primitive yo-yo apparatus (Figure 1), you replace the solid cylinder with a hollow cylinder of mass M, outer radius R, and inner radius R/2. Find the magnitude of the downward acceleration of the hollow cylinder. what is the Tension?
Physics
1 answer:
mr Goodwill [35]2 years ago
7 0

Answer:

Explanation:

Let T be the tension .

Applying newton's second law on the downward movement of the bucket

mg - T = ma

On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum

TR = Iα

TR = Ia/ R

T =  Ia/ R²

Replacing this value of T in the other equation

mg - T = ma

mg - Ia/ R²  = ma

mg =  Ia/ R² +ma

a ( I/ R² +m)= mg

a = mg / ( I/ R² +m)

mg - T = ma

mg - ma  = T

mg - m x mg / ( I/ R² +m) = T

mg - m²g / ( I/ R² +m ) = T

mg - mg / ( 1 + I / m R² ) = T

b ) T =  Ia/ R²

I = TR² / a

c ) Moment of inertia of hollow cylinder

I = 1/2  M ( R² - R² / 4 )

= 3/4 x 1/2 MR²

= 3/8 MR²

I / R² = 3/8 M

a = mg / ( I/ R² +m)

a = mg / ( 3/8 M + m )

T =  Ia/ R²

= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²

= \frac{3mMg}{(3M +8m)}

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d = 110 - 10 = 100 m

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v_f = 0

v_i = 20 m/s

now from kinematics equation we will have

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0 - 20^2 = 2(a)100

a = \frac{-400}{200} = -2 m/s^2

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3 0
2 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0
2 years ago
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Jobisdone [24]

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2 years ago
You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro
ehidna [41]

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The force applied to the spring is equal to the weight of the object suspended on it, so:

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The spring obeys Hook's law:

F=k\Delta x

where k is the spring constant and \Delta x is the stretching of the spring. Since we know \Delta x=2.92 cm=0.0292 m, we can re-arrange the equation to find the spring constant:

k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance \Delta x is equal to the elastic potential energy stored by the spring, given by:

W=U=\frac{1}{2}k\Delta x^2

Substituting k=907.5 N/m and \Delta x=8.80 cm=0.088 m, we find the amount of work that must be done:

W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J

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2 years ago
One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

a_{r} = 4\pi^{2}\cdot f^{2}\cdot R (1)

Where:

f - Frequency, measured in hertz.

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From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0
2 years ago
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