A 59.6 g box is left 17.5 cm from the centre of a merry-go-round. If the box has coefficients of friction of μs = 0.790 and μk = 0.550 with the merry-go-round, what is the maximum speed of the merry-go-round without having the box slide off of the merry-go-round?
To determine the maximum speed of the merry-go-round without having the box slide off, we need to consider the forces acting on the box.
1. Calculate the maximum static friction force (Fs):
Fs = μs * N
where μs is the coefficient of static friction and N is the normal force.
2. Calculate the normal force (N):
The normal force acting on the box is equal to its weight, which can be calculated as:
N = m * g
where m is the mass of the box and g is the acceleration due to gravity.
3. Calculate the maximum torque exerted by the box (τs):
τs = Fs * r
where r is the distance of the box from the center of the merry-go-round.
4. Calculate the maximum angular acceleration (α) of the merry-go-round:
α = τs / (I + m * r^2)
where I is the moment of inertia of the merry-go-round.
5. Calculate the maximum linear acceleration (a) of the merry-go-round:
a = α * r
6. Calculate the maximum velocity (v) of the merry-go-round:
v = √(a * r)
Let's plug in the given values:
Given:
- Mass of the box (m) = 59.6 g = 0.0596 kg
- Distance from the center (r) = 17.5 cm = 0.175 m
- Coefficient of static friction (μs) = 0.790
- Coefficient of kinetic friction (μk) = 0.550
We know that the maximum speed occurs when the static friction force is at its maximum value.
1. Calculate the maximum static friction force (Fs):
Fs = μs * N
2. Calculate the normal force (N):
N = m * g
3. Calculate the maximum torque exerted by the box (τs):
τs = Fs * r
4. Calculate the maximum angular acceleration (α) of the merry-go-round:
α = τs / (I + m * r^2)
5. Calculate the maximum linear acceleration (a) of the merry-go-round:
a = α * r
6. Calculate the maximum velocity (v) of the merry-go-round:
v = √(a * r)
Now let's calculate the values step by step:
Step 1: Calculate the maximum static friction force (Fs)
Fs = μs * N
Step 2: Calculate the normal force (N)
N = m * g
Step 3: Calculate the maximum torque exerted by the box (τs)
τs = Fs * r
Step 4: Calculate the maximum angular acceleration (α) of the merry-go-round
α = τs / (I + m * r^2)
Step 5: Calculate the maximum linear acceleration (a) of the merry-go-round
a = α * r
Step 6: Calculate the maximum velocity (v) of the merry-go-round
v = √(a * r)
Let's perform the calculations now.
To find the maximum speed of the merry-go-round without the box sliding off, we need to consider the forces acting on the box and the conditions for sliding.
Let's break down the problem step by step:
Step 1: Identify the forces acting on the box.
The main forces acting on the box are:
- The gravitational force pulling the box downwards (mg).
- The normal force exerted by the merry-go-round on the box (N).
- The static friction force (fs) when the box is not sliding.
- The kinetic friction force (fk) when the box is sliding.
Step 2: Determine the conditions for sliding.
The box will slide off the merry-go-round when the force of static friction (fs) cannot counteract the gravitational force pulling the box outward. The maximum force of static friction is given by fs = μsN, where μs is the coefficient of static friction. Therefore, for the box not to slide, we need fs to be greater than or equal to the gravitational force (mg).
Step 3: Calculate the maximum force of static friction.
The force of static friction can be calculated as fs = μsN. In this case, the normal force (N) is equal to the weight of the box, N = mg. So, fs = μsmg.
Step 4: Determine the centripetal force acting on the box.
When the merry-go-round spins, the box experiences a centripetal force that keeps it in circular motion. The centripetal force (Fc) acting on the box is given by Fc = (m)(v^2)/r, where m is the mass of the box, v is the velocity of the merry-go-round, and r is the distance from the center.
Step 5: Equate the force of static friction and the centripetal force.
To find the maximum speed, we need to equate fs and Fc, and solve for the velocity (v). Therefore, μsmg = (m)(v^2)/r.
Step 6: Solve for the maximum speed.
Rearrange the equation to solve for v:
v^2 = (μsmgr)/m
v = √(μsgr)
Now we can substitute the given values:
μs = 0.790 (coefficient of static friction)
g = 9.8 m/s^2 (acceleration due to gravity)
r = 17.5 cm = 0.175 m (distance from the center of the merry-go-round)
Plugging these values into the equation, we can calculate the maximum speed (v):
v = √(0.790 x 9.8 x 0.175) ≈ 2.565 m/s
Therefore, the maximum speed of the merry-go-round without the box sliding off is approximately 2.565 m/s.