A 3.00 kg ball is dropped from the roof of a building 176.4 m high. While the ball is falling to Earth, a horizontal wind exerts
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A)
B)
C)
D)
E)
Explanation:
A)
When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:
where
q is the charge of the particle (positive)
On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):
where
m is the mass of the particle
v is its final speed
According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:
And solving for v, we find the speed v at which the particle enters the cyclotron:
B)
When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is
where B is the strength of the magnetic field.
This force acts as centripetal force, so we can write:
where r is the radius of the orbit.
Since the two forces are equal, we can equate them:
And solving for r, we find the radius of the orbit:
(1)
C)
The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.
It can be calculated as the ratio between the length of the circumference () and the velocity of the particle (v):
(2)
From eq.(1), we can rewrite the velocity of the particle as
Substituting into(2), we can rewrite the period of revolution of the particle as:
And we see that this period is indepedent on the velocity.
D)
The angular frequency of a particle in circular motion is related to the period by the formula
(3)
where T is the period.
The period has been found in part C:
Therefore, substituting into (3), we find an expression for the angular frequency of motion:
And we see that also the angular frequency does not depend on the velocity.
E)
For this part, we use again the relationship found in part B:
which can be rewritten as
(4)
The kinetic energy of the particle is written as
So, from this we can find another expression for the velocity:
And substitutin into (4), we find:
So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.
Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).
The equilibrium conditions allow to find the results for the balance forces are:
- F₁ = 225.4 N
- F₂ = 558.6 N
When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.
∑ F = 0
∑ τ = 0
Where F are the forces and τ the torques.
The torque is the product of the force and the perpendicular distance to the point of support,
The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.
We write the translational equilibrium condition.
F₁ - W₁ - W₂ + F₂ = 0
We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.
F₂ 2 - W₁ 1 - W₂ 1.5 = 0
Let's calculate F₂
F₂ =
F₂ = (m g + M g 1.5)/ 2
F₂ =
F₂ = 558.6 N
We substitute in the translational equilibrium equation.
F₁ = W₁ + W₂ - F₂
F₁ = (m + M) g - F₂
F₁ = (12 +68) 9.8 - 558.6
F₁ = 225.4 N
In conclusion using the equilibrium conditions we can find the forces of the balance are:
- F₁ = 225.4 N
- F2 = 558.6 N
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