B.earth rotates on its axis in 24 hours; i.e., it rotates 15° of longitude per hour.
Answer:
72.75 kg m^2
Explanation:
initial angular velocity, ω = 35 rpm
final angular velocity, ω' = 19 rpm
mass of child, m = 15.5 kg
distance from the centre, d = 1.55 m
Let the moment of inertia of the merry go round is I.
Use the concept of conservation of angular momentum
I ω = I' ω'
where I' be the moment of inertia of merry go round and child
I x 35 = ( I + md^2) ω'
I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19
35 I = 19 I + 1164
16 I = 1164
I = 72.75 kg m^2
Thus, the moment of inertia of the merry go round is 72.75 kg m^2.
Answer:
a. 12.12°
b. 412.04 N
Explanation:
Along vertical axis, the equation can be written as
T_1 sin14 + T_2sinA = mg
T_2sinA = mg - T_1sin12.5 ....................... (a)
Along horizontal axis, the equation can be written as
T_2×cosA = T_1×cos12.5 ......................... (b)
(a)/(b) given us
Tan A = (mg - T_1sin12.5) / T_1 cos12.5
= (176 - 413sin12.5) / 413×cos12.5
A = 12.12 °
(b) T2 cosA = T1 cos12.5
T2 = 413cos12.5/cos12.12
= 412.04 N
Answer:
the correct one is b
the difference between the final moment and the initial moment
Explanation:
The momentum is related to the moment
I = ΔP
∫ F dt = p_f - p₀
where p_f and p₀ are the final and initial moments, respectively
When checking the different answers, the correct one is b
the difference between the final moment and the initial moment
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, 
rev/sec
rad/sec
= 5.89 rad/sec
Therefore, force required,
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
