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Andrews [41]
3 years ago
5

The shaft of radius c is subjected to a distributed torque t, measured as torque/unit length of shaft. Shaft A B of length L, fi

xed at B, with x measured from B. Torque t = t sub 0 [1 + (x over L) squared], from t sub 0 at B to 2 t sub 0 at A. The shear modulus is G. Part A Determine the angle of twist at end A. Express your answer as an expression in terms of the variables t0, L, c, and G and any necessary constants.

Physics
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

Angle of twist;Φ = (to•L²)/Gπc⁴

Explanation:

I have attached the explanation.

However, I used D for radius of shaft against c used in the question. So, if we replace the D with c, we'll have;

Φ = (to•L²)/Gπc⁴

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A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1.
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Answer:98

Explanation:hope this helps!

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What is the average velocity of a car that travels 50 kilometers due west in 0.50 hour?
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Average Velocity= displacement/time   Av=50/0.50   Av=100

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A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 106 meters, find the mass of
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7.36 × 10^22 kg

Explanation:

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Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
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