Rotational motion solved problems
Notes about calculating rotational motion
▲
When dealing with circular motion there are some parameters that we should be familiar with.
T
Time of 1 cycle
s
f
frequency
Hz
ω
angular velocity
rad / sec
α
angular acceleration
rad / sec
^{2}
v
tangential velocity
m / s
a
_{R}
radial acceleration
m / s
^{2}
F
_{R}
Centripetal force
N
r
radius of motion
r
m
If angular velocity is given in
(
rpm
)
then ω =
(
rpm
)
2π / 60
[
rad / s
]
Forces conversion: 1
[
kgf
]
= 9.8
[
N
]
= 9.8
[
kgm · m / s
^{2}
]
Rotationsl motion example - 1
▲
Find the angle α if the mass m is rotating at an angular velocity of ω around the vertical rode, the rope length is L.
From centrifugal force equation we have:
T sinα − m ω
^{2}
R = 0
(1)
From forces equilibrium in the y direction:
T cosα − m g = 0
(2)
We have also the equation:
R = L sinα
(3)
Substitute (1) into (2) to get the value of angle α
If we substitute equation (3) into (1) and (2) we get:
From the equation above we notice that the angle α is not a function of the mass m.
The height of the pendulum is h = L cosα if we substitute the value of cosα we get: h = g / ω
^{2}
The tension in the chord is calculated by eq. (1) and (3) :
T = m ω
^{2}
L
Rotationsl motion example - 2
▲
The mass m is turning vertically at a constant angular velocity of ω rad/s if the rope length is R, find the tension in the rope at points A, B and C.
From the forces diagram we can write the total forces acting in the radial (to the center):
In the R direction:
T + m g sinα − m ω
^{2}
R = 0
(1)
And the tension T is
T = m ω
^{2}
R − m g sinα
And the tension at the points A, B and C are found by inserting the angles into the expression for T.
At point A (α = − 90 °)
T = m (ω
^{2}
R + g)
At point C (α = 90 °)
T = m (ω
^{2}
R − g)
At point B
T = m (ω
^{2}
R − g sinα)
We can see that the tension in the rope is minimum at point C and maximum at point A, and between this points T is a function of the angle α .
At the top the tension in the rope has its minimum value therfore in order to keep the mass not to fall the angular velocity should be at least:
m ω
^{2}
R = m g
→
so the angular velocity should be:
If we substitute the velocity v = ω R , we obtaine the minimum speed:
Rotationsl motion example - 3
▲
The mass m is released from rest at point A. When the mass reaches vertical position, the rope turns around a nail which is located at a distance of r from the end of the rope. Find the angle α that will cause a complete rotation of the mass around the nail.
We will take the axes so that the x direction is pointing to the center of the rotation and the y direction is always tangential to the circle in this axes system the equations are more simple:
The velocity of the mass at the vertical position when the rope reches the nail is calculated by energy methods.
The initial potantial energy is converted to kinetic energy:
Falling distance is:
h = R − R cosα = R (1 − cosα)
The velocity at point B is:
(1)
In order to complete a hole rotation around the nail the kinetic energy at point B (vertical position) should be equal to the potential energy at height 2r (upper position of the mass after swirling around the nail) plus the kinetic energy at top due to the velocity v
_{T}
.
Energy balance:
(2)
From free body diagram
(3)
From eq. (3)
v
_{T}
^{2}
= g r
(4)
Substitute eq. (1) and (4) into (2) we get:
m 2 g h / 2 = m g 2 r + m g r / 2
h = 5 r / 2
Substitut the value of h to the privious found value of h we get:
If point A is horizontaly located relative to point O and the ball released from this position, find the maximum value of r that will cause a complete rotation around the nail.
In this case:
h = R
and from eq. (1) v
_{B}
^{2}
= 2 g R substitute this values into eq. (2) we get:
And finally:
r = 2 R / 5
Pendulum on cart example - 4
▲
A pendulum is tilted to an angle θ due to horizontal acceleration a [m / s
^{2}
] of the cart, the mass of the pendulum is m, find the angle θ and the tension in the cord expressed by the mass and acceleration.
From the free body diagram the forces equilibrium in x and y direction are:
x direction:
T sinθ = m a
(1)
y direction:
T cosθ = m g
(2)
By dividing eq. (1) with eq. (2) we can find the angle θ :
(3)
From equation (3) we see that the tilt angle is a function of the acceleration a and the gravitational acceleration g.
The tension in the rope can also be eliminated:
or
Rotationsl motion example - 5
▲
A bullet is shoot into a block of wood suspended from a light flexible wire. The bullet lodges in the wood and causes it to tilt to an angle θ. Find the angle θ as a function of m, M, L, v
_{1}
and v
_{2}
.
Note: We suppose that the impact is ideal and there are no heat and plastic looses.
After impact the velocity of the bullet and wood are the same, from conservation of momentum we have:
m v
_{1}
= (m + M) v
_{2}
(1)
From conservation of energy:
1/2 (m + M) v
_{2}
^{2}
= (m + M) g L cosθ
(2)
From equation
(2)
eliminating θ we get:
From eq. (1) we have:
If we have to find v
_{1}
then from the equations
(1)
we have:
And v
_{1}
is:
(3)
For example if a bullet mass is 10 gram and the wood mass is 1 kg. After the collision it is observed that the wood mass center is 20 cm above the original height. Find the velocity of the bullet before impact.
From eq. (3) we have:
Rotationsl motion example - 6
▲
Two sphers each with mass m, are connected by four ropes of length L as shown in the figure, both masses are connected to a mass M that can move freely up and down the rode. The system is rotating at an angular velocity ω rad / s. Find the angle θ as a function of m, M, L and ω.
Note: If the weight W is given in kg then m = W / g [kgm]
From forces equilibrium equations of the spheres we have:
x direction
F
_{1}
cosα + F
_{2}
cosα − m r ω
^{2}
= 0
(1)
y direction
F
_{1}
sinα − F
_{2}
sinα − m g = 0
(2)
From forces equilibrium equation of the mass M in y direction.
2F
_{2}
cosθ − M g = 0
(3)
Now we substitute the values:
r = L sinθ and α = 90 − θ
From eq.
(1)
we get:
F
_{1}
+ F
_{2}
= m L ω
^{2}
From eq.
(2)
we get:
Solving the above equations for F
_{1}
and F
_{2}
we get:
Substitute F
_{2}
into (3) we get:
→
Find the angular velocity if the angle θ is given:
From eq. (3) we have:
Inserting this value to the value of F
_{2}
we get:
Motorist turn example - 7
▲
A motorist is driving at a velocity of v [m / s], he is performing a turn of a radius r. Find the angle θ that the motorcycle should be during the turn, and the minimum friction required in order to prevent the slipping of the motorcycle.
During the turn the centrifugal force is balanced by the weight of the motorcycle, from equilibrium of forces we get:
x direction:
m ω
^{2}
r − f = 0
(1)
y direction:
N − m g = 0
(2)
Moments at point A:
m g sinθ L − m ω
^{2}
r cosθ L = 0
(3)
The value L is the distance of the center of mass to the wheel end (point A), but L has no effect on the calculations because after dividing eq. (3) by L it disappears.
From eq. (3) we can find the tilt angle θ :
→
The friction force is:
f = μ
_{s}
N this force is balanced by the centripetal force.
Substitute the value of f into eq. (1) and (2) we get:
(4)
In order not to slip the friction coefficient should be bigger then the value calculated:
Rotationsl motion example - 8
▲
A small ball is in rest at point A. the ball is then slipping down on a circular path of radius R. Find the height h of the point B where the ball separates from the circular path.
Note: we suppose that the radius of the ball is much less then the radius of the circular path and there is no friction during the motion.
When separation begines the increasing centrifugal force due to gravitational acceleration g is balanced by the value of mg cosθ. We choose the y axis pointing to the center of the circle path.
→
(1)
In order to find the velocity at B we use energy method.
→
(2)
The value of h expressed by R is:
h = R − R cosθ = R (1 − cosθ)
(3)
Substitute (2) and (3) into eq. (1) to get the separation height:
(4)
And the speed at separation point is:
Seperation angle from eq. (3) is:
Rotationsl motion example - 9
▲
A small ball is released from rest at point A, at point B it enters a circular tube of radius r. Find the height h that the ball should be released in order to reach point C, what is the velocity of the ball at point C suppose that the friction is negligible.
At point C the minimum velocity should be:
→
(1)
The energy of the ball at points B compared to C is:
→
(2)
Now we can compare the energy of the ball at points A and B.
(3)
Isolating h and substitute equation (2) into eq. (3) we get the released height:
(4)
If high h is equal to 4r what is the force of the ball on the circles surface at points B and C.
From energy balance at points A and B we have:
The reaction force at point B is:
The velocity at point C from energy:
The reaction force at point C is:
Rotationsl motion example - 10
▲
A circular tube is rotating around the vertical axis at a fixed angle of θ degree and an angular velocity ω [rad / s]. Inside the frictionless tube a ball is released, if the mass of the ball is m, determine the radius r that the ball will reach.
We choose the x-y axis to coincide with the tube axis. The forces along y axis that causes the ball
to be at equilibrium are:
m r ω
^{2}
sinθ = m g cosθ
(1)
Isolate r from equation (1) we get:
If friction coefficient between the mass and tube wall is μ
_{s}
and the mass is located at a distance r from the rode find the range of the angular velocity that will keep the mass at rest.
To find the maximum anguler velocity the direction of the mass is upward and the friction force is downward.
Forces in x direction:
N = m ω
^{2}
r cosθ + m g sinθ
(1)
Forces in y direction:
m g cosθ + f = m ω
^{2}
r sinθ
(2)
The friction force is f = N μ
_{s}
Maximum angular velocity is:
To find the minimum anguler velocity the direction of the mass is downward hence friction force is upward.
Forces in x direction:
N = m ω
^{2}
r cosθ + m g sinθ
(3)
Forces in y direction:
m g cosθ − f = m ω
^{2}
r sinθ
(4)
Minimum angular velocity is:
Road banking example - 11
▲
v
m/s
θ
deg
R
m
μ
M
kgm
f
N
N
N
Input:
Icy road friction μ
_{s}
= 0
Flate road θ = 0
A car is traveling at a speed of v m/s when it enters a turn of radius R the friction of the wheels with the road is μ
_{s}
in order to prevent skidding of the car the road is banked by an angle of θ degree. Find the maximum speed with no skidding of the car.
The maximum friction force of the car is:
f = N μ
_{s}
For present solution we neglect the moment caused by the friction of the wheels compared to the center of the car mass.
The axes is chose so that the x is along the road surface to the center and y is normal to the road surface.
Forces along x axis:
(1)
Forces along y axis:
(2)
Isolate v from eq. (1) and (2) we get:
If we need to find the banking angle then from eq. (1) and (2) we have:
Note that the velocity v and banking angle is not a function of the mass.
If Rgμ
_{s}
is bigger then v
^{2}
then the angle is negative that means that no banking of the road is needed.
from (1) and (2) μ
_{s}
is:
Radius of road curvature: