Answer:
I think it's claim 1 because the same mass would cause them both to either fall off or to stay there
Answer:
Note: Angular momentum is always conserved in a collision.
The initial angular momentum of the system is
L = ( It ) ( ωi )
where It = moment of inertia of the rotating circular disc,
ωi = angular velocity of the rotating circular disc
The final angular momentum is
L = ( It + Ir ) ( ωf )
where ωf is the final angular velocity of the system.
Since the two angular momenta are equal, we see that
( It ) ( ωi ) = ( It + Ir ) ( ωf )
so making ωf the subject of the formula
ωf = [ ( It ) / ( It + Ir ) ] ωi
Explanation:
Answer:
6 m/s.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 4 m/s
Acceleration (a) = 2 m/s²
Distance (s) = 5 m
Final velocity (v) =?
The final velocity of the object can obtained as shown below:
v² = u² + 2as
v² = 4² + (2 × 2 × 5)
v² = 16 + 20
v² = 36
Take the square root of both side.
v = √36
v = 6 m/s
Therefore, the final speed of the object is 6 m/s.
Answer:
(A) 0.54 kg.m^{2}
(B) 0.0156 N
Explanation:
from the question you would notice that there are some missing details, using search engines you can find similar questions online here 'https://www.chegg.com/homework-help/questions-and-answers/small-ball-mass-120-kg-mounted-one-end-rod-0860-m-long-negligible-mass-system-rotates-hori-q7245149'
here is the complete question:
A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of 2.60 x 10^{-2} N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?.
solution
mass of the ball (m) = 1.5 kg
length of the rod (L) = 0.6 m
angular velocity (ω) = 4900 rpm
air drag (F) = 2.60 x 10^{-2} N = 0.026 N
(take note that values from the original question are used, with the exception of the air drag which was not in the original question)
(A) because the rod is mass less, the rotational inertia of the system is the rotational inertia of the rod about the other end, hence rotational inertia =
where m = mass of ball and L = length of rod
=
= 0.54 kg.m^{2}
(B) The torque that must be applied to keep the ball in motion at constant speed = FLsin90
= 0.026 x 0.6 x sin 90 = 0.0156 N
Answer:
Explanation:
The force due to gravitation is equal to zero for each of the masses.
M1= 181kg
M2= 712kg
m = 72.6kg
The distance between M1 and M2 is said to be fixed , therefore no value should be given I.e it's a constant.
From the formula for gravitational force we have that
F = GMm/r^2
GmM1/(d-r)^2. = GmM2/r^2
where r is the distance between the 72.6 kg and 712kg
d is the distance between M1 and M2
Solving mathematically
r(√M1+√M2) = d√M2
r = d√M2/√M1 + √M2
d×26.68/ 13.45+26.68
d×26.68/40.13
r = 0.665d