Answer:

Explanation:
The equation for kappa ( κ) is

we can find the maximum of kappa for a given value of b using derivation.
As b is fixed, we can use kappa as a function of a

Now, the conditions to find a maximum at
are:


Taking the first derivative:








This clearly will be zero when

as both are greater (or equal) than zero, this implies

The second derivative is




We dcan skip solving the equation noting that, if a=b, then

at this point, this give us only the first term

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum
the value of kappa is



Answer:
v (minimum speed) = 2.90 m/sec.

Maximum value of speed will occur at lowest point of vertical circle.
Explanation:
a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?
Using the force balance expression at the top of the circle,
Gravitational Force + Tension force = Centrifugal force

Given that : T = 0
R = length of string = 0.86 m
mass of the spinning rock = 0.75 kg


v (minimum speed) = 2.90 m/sec.
b) what is the maximum speed the rock can have so that the string does not break?
Here the force balance at bottom of circle is represented by the illustration:

Given that:
maximum tension T = 45 N
maximum speed v = ??
mass m = 0.75 kg
∴

c)
At what point in the vertical circle does this maximum value occur?
Maximum value of speed will occur at lowest point of vertical circle.
This is so because at the lowest point; the tension in string will be maximum.
Centripetal force is given by F= mv²/r.
Given: m = 0.5 kg, v = 3 m/s, r = 0.5 m
Putting values,
F= mv²/r = 0.5× 3²/0.5 = 9 N
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.