At a playground, a 20-kg child sits on a spinning merry-go-round, as shown from above in (Figure 1) . The merry-go-round completes one revolution every 6.2 s, and the child sits at a radius of r=1.8m.
a. What is the force of static friction acting on the child?
b.What is the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping?
a. Well, since the child is sitting on the merry-go-round, we know that there must be a force of static friction acting on the child. Let's call it "Fsf" for short. Now, we can use the formula for centripetal force to find the value of Fsf. The centripetal force is given by the equation Fc = mω^2r, where m is the mass of the child, ω is the angular velocity of the merry-go-round, and r is the radius of the child's position on the merry-go-round.
First, let's find the angular velocity. The merry-go-round completes one revolution every 6.2 s, which means it completes 2π radians (one revolution) in that time. So, ω = 2π/6.2 s.
Now, let's plug in the values we know into the formula and solve for Fsf:
Fc = mω^2r
Fsf = mω^2r
Plugging in the values:
Fsf = (20 kg)([(2π/6.2 s)^2] x 1.8 m)
Calculating that gets us:
Fsf ≈ 174.98 N
So, the force of static friction acting on the child is approximately 174.98 N.
b. Now, let's move on to finding the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping. The coefficient of static friction is represented by the symbol "μs".
To prevent slipping, the force of static friction Fsf must be equal to or greater than the maximum frictional force Ffmax. The maximum frictional force can be calculated using the formula Ffmax = μsN, where N is the normal force.
In this case, the normal force N is equal to the weight of the child, N = mg, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, let's plug in the values for Fsf and N into the formula and solve for μs:
Fsf = Ffmax
μsN = Ffmax
μsmg = Ffmax
Plugging in the values:
μs(20 kg)(9.8 m/s^2) = 174.98 N
Solving for μs:
μs ≈ 0.898
So, the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping is approximately 0.898.
To find the force of static friction acting on the child, we need to consider the circular motion of the child on the merry-go-round.
a. The centripetal force is provided by the static friction between the child and the merry-go-round. This force can be calculated using the formula:
F = (m * v^2) / r,
where:
F is the centripetal force,
m is the mass of the child (20 kg),
v is the linear velocity of the child, and
r is the radius of the circular path (1.8 m).
First, we need to find the linear velocity of the child. Since the merry-go-round completes one revolution every 6.2 s, the velocity can be calculated as follows:
v = circumference / time,
v = (2 * π * r) / t,
v = (2 * 3.14 * 1.8) / 6.2,
v ≈ 1.836 m/s.
Now, we can substitute the values into the formula to find the centripetal force:
F = (m * v^2) / r,
F = (20 * 1.836^2) / 1.8,
F ≈ 37.55 N.
Therefore, the force of static friction acting on the child is approximately 37.55 N.
b. To find the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping, we can use the following formula:
μs = Ff / Fn,
where:
μs is the coefficient of static friction,
Ff is the force of static friction, and
Fn is the normal force.
The normal force is equal to the weight of the child, which is given by:
Fn = m * g,
Fn = 20 kg * 9.8 m/s^2,
Fn = 196 N.
Therefore, we can substitute the values into the formula to find the coefficient of static friction:
μs = Ff / Fn,
μs = 37.55 N / 196 N,
μs ≈ 0.191.
Therefore, the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping is approximately 0.191.
To find the force of static friction acting on the child, we need to consider the centripetal force acting on the child when the merry-go-round is spinning. The centripetal force is provided by the static friction between the child and the merry-go-round.
a. To calculate the force of static friction, we can use the formula for centripetal force:
Fc = m * ac
Where Fc is the centripetal force, m is the mass of the child, and ac is the centripetal acceleration.
First, let's find the centripetal acceleration. Since the merry-go-round completes one revolution every 6.2 seconds, we can determine the angular velocity (ω) using the formula:
ω = 2π / T
Where T is the time period for one revolution.
ω = 2π / 6.2 s
Now, using the angular velocity, we can find the centripetal acceleration (ac) using the formula:
ac = ω² * r
Where r is the radius of the child's position on the merry-go-round.
ac = (ω²) * r
= (2π / 6.2 s)² * 1.8 m
Next, we can substitute the centripetal acceleration and the mass of the child into the formula for centripetal force:
Fc = m * ac
= 20 kg * [(2π / 6.2 s)² * 1.8 m]
This will give us the force of static friction acting on the child.
b. To find the minimum coefficient of static friction between the child and the merry-go-round to keep the child from slipping, we can use the equation:
μs = (Fs) / (N)
Where μs is the coefficient of static friction, Fs is the force of static friction, and N is the normal force.
The normal force (N) is equal to the weight of the child (mg), where g is the acceleration due to gravity (approximately 9.8 m/s²).
N = mg
= 20 kg * 9.8 m/s²
Now, substitute the force of static friction and the normal force into the formula for the coefficient of static friction:
μs = (Fs) / (N)
= (Force of static friction) / (20 kg * 9.8 m/s²)
This will give us the minimum coefficient of static friction needed to keep the child from slipping on the merry-go-round.
F = m v^2/r = m omega^2 r
m = 20
v = 2 pi r/6.2 = 2 pi(1.8/6.2)
r = 6.2
so solve for F, the mass times centripetal a
mu = Ac/9.81 = (v^2/r) /9.81