Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th
2 answers:
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Answer:
The frictional force needed to overcome the cart is 4.83N
Explanation:
The frictional force can be obtained using the following formula:
where is the coefficient of friction = 0.02
R = Normal reaction of the load = =
=
Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.
F = 4.83 N
Hence, the frictional force needed to overcome the cart is 4.83N
Answer:
The angular speed after 6s is .
Explanation:
The equation
relates the moment of inertia of a rigid body, and its angular acceleration
, with the force applied
at a distance
from the axis of rotation.
In our case, the force applied is , at a distance
, to a ring with the moment of inertia of
; therefore, the angular acceleration is
Therefore, the angular speed which is
after 6 seconds is
Answer:
The torque on the child is now the same, τ.
Explanation:
- It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
- In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
- The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
- When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:
- So, τ = 3/2*m*r²*α (2)
- When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:
- Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:
τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.