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Bingel [31]
3 years ago
13

A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation.

Physics
1 answer:
dezoksy [38]3 years ago
4 0

Hi there!

We can use the conservation of angular momentum to solve.

L_i = L_f\\\\I\omega_i = I\omega_f

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

\text{Solid cylinder:} I = \frac{1}{2}MR^2\\\\\text{Object around center:} = MR^2

Begin by converting rev/sec to rad sec:


\frac{0.17rev}{s} * \frac{2\pi rad}{1 rev} = 1.068 \frac{rad}{s}

According to the above and the given information, we can write an equation and solve for ωf.

1.068(\frac{1}{2}(34)(1.6)^2 + (79)(1.6)^2) = \omega_f(\frac{1}{2}(34)(1.6^2) + 79(0^2))\\\\\omega_f = \boxed{6.03 \frac{rad}{sec}}

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If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the
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Either 175 N or 157 N depending upon how the value of 48° was measured from.    
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.    
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.    
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".     
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:  
U = F*cos(48)  
where  
U = Useful force  
F = Force applied    
So solving for F and calculating gives:  
U = F*cos(48)  
U/cos(48) = F  
117 N/0.669130606 = F  
174.8537563 N = F    
So 175 Newtons of force is required in this situation.    
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get  
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3 years ago
A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very fa
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To solve this problem we will apply the concepts related to the electric field, linear charge density and electrostatic force.

The electric field is

E = \frac{\lambda}{2\pi \epsilon_0 r}

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\lambda= Linear charge density

\epsilon_0 = Permittivity of free space

r = Distance

The linear charge density can be written as,

Linear charge density is given as

\lambda = \frac{q}{L}

Replacing,

E = \frac{\frac{q}{L}}{2\pi \epsilon_0 r}

E = \frac{q}{2\pi \epsilon_0 rL}

The initial and final electric Force can be written as function of the charge and the electric field as

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F_f = E_f q

If we replace the value for the electric field we have,

F_i = (\frac{q}{2\pi \epsilon_0 rL})q = (\frac{q^2}{2\pi \epsilon_0 rL})

Length is one third at the end, then

F_f = (\frac{q}{2\pi \epsilon_0 r(L/3)})q = (\frac{3q^2}{2\pi \epsilon_0 rL})

The ratio of the force is

\frac{F_f}{F_i} = \frac{(\frac{3q^2}{2\pi \epsilon_0 rL})}{(\frac{q^2}{2\pi \epsilon_0 rL})}

\frac{F_f}{F_i} = 3

Therefore the required ratio is 3

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