Answer: hello some part of your question is missing attached below is the missing detail
answer :
<em>w</em>f = M( v cos∅ )D / I
Explanation:
The Angular speed <em>wf </em>of the system after collision in terms of the system parameters and I can be expressed as
considering angular momentum conservation
Li = Lf
M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) <em>w</em>f
where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )
In terms of system parameters and I
<em>w</em>f = M( v cos∅ )D / I
Answer:
α=9.42 rad/s T= 0.014 rad/s
Explanation:
Low speed= Wi=3 rps = 3*2*π rad/s= 6π rad/sec
Medium speed= Wf= 9 rps= 9*2*π rad/sec= 18 rad/sec
time= t= 4 sec
Angular acceleration=α=
⇒α=(18 π - 6 π)/4
⇒α=9.42 rad/s
b).
Rotational inertia= I = 1.5*10^-3 kg-m^2
Net torque= T = I*α = (1.5*10^-3) * (9.42) rad/s^2
⇒T= 0.014 rad/s^2
Answer:
The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²
The rocket's motion for analysis sake is divided into two phases.
Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m
Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.
Explanation:
The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.
The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.
The detailed step by step solution to the problems can be found in the attachment below.
Thank you and I hope this solution is helpful to you. Good luck.
