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marishachu [46]
3 years ago
14

Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec

t is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.84 Hz?
Physics
1 answer:
yKpoI14uk [10]3 years ago
8 0

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

mg=kx

k=\dfrac{mg}{x}

k=\dfrac{3.74\times 9.8}{0.0161}

k = 2276.52 N/m

The frequency of vibration is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

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A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

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The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th
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Answer:

The force exerted by the biceps is 143.8 kgf.

Explanation:

To calculate the force exerted by the biceps, we calculate the momentum in the elbow.

This momentum has to be zero so that her forearm remains motionless.

Being:

W: mass weight (6.15 kg)

d_W= distance to the mass weight (0.425 m)

A: weight of the forearm (2.25 kg)

d_A: distance to the center of mass of the forearm (0.425/2=0.2125 m)

H: force exerted by the biceps

d_H: distance to the point of connection of the biceps (0.0215 m)

The momemtum is:

H*d_H-A*d_A-W*d_W=0\\\\H=(A*d_A+W*d_W)/d_H\\\\H=(2.25*0.2125+6.15*0.425)/0.0215\\\\H=(0.478125+2.61375)/0.0215\\\\H=3.091875/0.0215=143.8

The force exerted by the biceps is 143.8 kgf.

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