Explanation:
Given parameters:
Frequency = 500Hz
Unknown:
a. Period of rotation
b. Time required to complete 7 rotation
Solution:
The period of rotation is the time taken to complete a turn.
P =
P is the period
t is the time
n is the number of turns
Period is the inverse of frequency;
P =
= 0.002s
b.
Time required to complete 7 rotation;
P =
t = P x n
t is the time
P is the period
n is the number of rotations
t = 0.002 x 7 = 0.014s
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
![F_{frmax} = \mu_{s} *F_{n} (1)](https://tex.z-dn.net/?f=F_%7Bfrmax%7D%20%3D%20%5Cmu_%7Bs%7D%20%2AF_%7Bn%7D%20%281%29)
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
![F_{c} = m* \omega^{2} * r (2)](https://tex.z-dn.net/?f=F_%7Bc%7D%20%3D%20%20m%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%282%29)
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
![F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)](https://tex.z-dn.net/?f=F_%7Bfrmax%7D%20%3D%20m%2A%20%5Cmu_%7Bs%7D%20%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%283%29)
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
![m* g = m* \mu_{smin} * \omega^{2} * r (4)](https://tex.z-dn.net/?f=m%2A%20g%20%3D%20m%2A%20%5Cmu_%7Bsmin%7D%20%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%284%29)
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
![g = \mu_{smin} * \omega^{2} * r (5)](https://tex.z-dn.net/?f=g%20%3D%20%5Cmu_%7Bsmin%7D%20%2A%20%5Comega%5E%7B2%7D%20%2A%20r%20%285%29)
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
![60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)](https://tex.z-dn.net/?f=60%20rev%2Fmin%20%2A%20%5Cfrac%7B2%2A%5Cpi%20rad%7D%7B1%20rev%7D%20%2A%5Cfrac%7B1min%7D%7B60%20sec%7D%20%3D6.28%20rad%2Fsec%20%286%29)
- Replacing by the givens in (5), we can solve for μsmín, as follows:
![\mu_{smin} = \frac{g}{\omega^{2} *r} = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)](https://tex.z-dn.net/?f=%5Cmu_%7Bsmin%7D%20%3D%20%5Cfrac%7Bg%7D%7B%5Comega%5E%7B2%7D%20%2Ar%7D%20%20%3D%20%5Cfrac%7B9.8m%2Fs2%7D%7B%286.28rad%2Fsec%29%5E%7B2%7D%20%2A2.5%20m%7D%20%3D0.1%20%287%29)
Answer:
Explanation:
Since the wires attract each other , the direction of current will be same in both the wires .
Let I be current in wire which is along x - axis
force of attraction per unit length between the two current carrying wire is given by
x ![\frac{2 I_1\times I_2}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20I_1%5Ctimes%20I_2%7D%7Bd%7D)
where I₁ and I₂ are currents in the wires and d is distance between the two
Putting the given values
285 x 10⁻⁶ = 10⁻⁷ x ![\frac{2\times25.5\times I_2}{.3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Ctimes25.5%5Ctimes%20I_2%7D%7B.3%7D)
I₂ = 16.76 A
Current in the wire along x axis is 16.76 A
To find point where magnetic field is zero due the these wires
The point will lie between the two wires as current is in the same direction.
Let at y = y , the neutral point lies
k 2 x
= k 2 x ![\frac{25.5}{.3-y}](https://tex.z-dn.net/?f=%5Cfrac%7B25.5%7D%7B.3-y%7D)
25.5y = 16.76 x .3 - 16.76y
42.26 y = 5.028
y = .119
= .12 m
Answer:
F= MASS *ACCELERATION
12=6*a
a=2 m/sec^2
v^2-u^2=2*a*s
v^2=2*2*3
v=sqrt(12)
v=3.5 m/sec
work done by force= change in kinetic energy
F*S=0.5*M*U^2-0.5*M*V^2
12*3=0 0.5*6*V^2
V=3.46 M/SEC
Explanation: