Respuesta :
Answer:
option A
Explanation:
given,
distance of merry-go-round = 1.5 m
coefficient of static friction
acceleration due to gravity = 9.81 m/s²
1. friction
F = μ mg
2. angular force
[tex]F= \dfrac{mv^2}{R}[/tex]
to find maximum v, find at what v both forces are equal;
[tex]\mu mg = \dfrac{mv^2}{R}[/tex]
[tex]v^2=R \mu g[/tex]
[tex]v = \sqrt{1.5\times 0.7\times 9.81}[/tex]
v = 3.21 m/s
to find angular speed,
ω =[tex]\dfrac{v}{R}[/tex]
ω =[tex]\dfrac{3.21}{1.5}[/tex]
ω = 2.14
so, the correct answer is option A
Centrifugal forces act outwards from the center. The maximum constant angular speed of the merry-go-round for which the child will not start to slide is 2.14 rad/s.
What is centrifugal force?
Centrifugal force can be defined as the outward force that is applied to an object of mass m when it is rotated about an axis. It is given by the formula,
[tex]F = \dfrac{mv^2}{R} = m\omega^2R[/tex]
Given to us
Radius, r = 1.50 m
Coefficient of static friction, μ = 0.7
Acceleration due to gravity, g = 9.81 m/s²
If we draw the free body diagram of the child, we will see 4 forces are applied to the child,
- The normal force which is upwards, N
- Weight of the child which is downwards, W
- The friction force will try to keep the kid on the merry-go-round, therefore, towards the inside. F
- Centrifugal force towards the outside, [tex]F_c[/tex]
If we calculate all the vertical forces,
[tex]\sum F_y = 0\\N = W\\N = mg\\[/tex]
If we calculate all the horizontal forces,
[tex]\sum F_x = 0\\\\F = F_c\\\\\mu\ N = m\times \omega^2 R\\\\\mu\ mg = m\times \omega^2 R\\\\\mu\ g = \omega^2 R\\\\\omega = \sqrt{\dfrac{\mu g}{R}}\\\\\omega = 2.14\rm \ red/sec^[/tex]
Hence, the maximum constant angular speed of the merry-go-round for which the child will not start to slide is 2.14 rad/s.
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