A 10 kg cart is attached to a rope that forces the cart to travel along a circular path. The rope has a length of 2.7 meters and
1 answer:
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The moment of inertia is
Explanation:
The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.
The moment of inertia of the rod about its centre is given by
where
M = 24 kg is the mass of the rod
L = 0.96 m is the length of the rod
Substituting,
The moment of inertia of one ball is given by
where
m = 50 kg is the mass of the ball
is the distance of each ball from the axis of rotation
So we have
Therefore, the total moment of inertia of the system is
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Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) =
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
