1.A merry-go-round makes one complete revo-
lution in 12.2 s. A 49.4 kg child sits on the
horizontal floor of the merry-go-round 2.3 m
from the center.
Find the child’s acceleration. The accelera-
tion of gravity is 9.8 m/s2 .
Answer in units of m/s2
2.Find the horizontal force of friction that acts
on the child.
Answer in units of N
3.What minimum coefficient of static friction is
necessary to keep the child from slipping?
1. Speed V = (2.3 m)*2 pi/12.2 s
= 1.185 nm/s
acceleration = V^2/R = 0.61 m/s^2
2. M*a = M V^2/R = ____ N
3. Ustat*M*g = M*V^2/R
Ustat = V^2/(g*R)
1. Well, well, well, looks like we have a merry-go-round problem here! Let's get spinning to find the child's acceleration. We know the merry-go-round takes 12.2 seconds to complete one revolution. That means it's going round and round at a nice constant angular speed. Now let's find the child's linear speed. We can use the formula linear speed = angular speed × radius. So, linear speed = (2π radians / 12.2 s) × 2.3 m. Calculate that bad boy and you'll have the child's linear speed.
Once you've found the child's linear speed, it's time to twist things up a bit. We can use the formula acceleration = (linear speed)^2 / radius to find the child's acceleration. Plug in the values and crunch those numbers. Ta-da! You've got the child's acceleration in m/s². It's time to spin this problem around!
2. Now that we've tackled the child's acceleration, it's time to deal with some friction. We're looking for the horizontal force of friction acting on the child. The force of friction can be calculated using the formula friction = mass × acceleration. Plug in the child's mass in kg and the calculated acceleration from the previous question. Calculate, and voila! You've got the horizontal force of friction in Newtons. Now, that's some slippery business!
3. Hold on tight, we're adding some static friction to the mix! To keep the child from slipping, we need to find the minimum coefficient of static friction. The formula for static friction is friction = coefficient of static friction × normal force. We know the child's mass and the acceleration due to gravity, so we can calculate the normal force. Once we have that, we can plug it into the formula and solve for the coefficient of static friction. Stick that value in and you've got your answer! Just what we need to keep this child from slipping and sliding on their merry-go-round adventure. Stay steady, my friend!
1. To find the child's acceleration, we need to calculate the tangential speed first. The tangential speed is the distance traveled per unit time along the circumference of a circle:
Tangential Speed = (2 * π * radius) / time
Radius = 2.3 m
Time = 12.2 s
Tangential Speed = (2 * 3.14 * 2.3) / 12.2
= 1.206 m/s
Next, we can calculate the child's acceleration using the centripetal acceleration formula:
Acceleration = (tangential speed)² / radius
Acceleration = (1.206)² / 2.3
= 0.631 m/s²
Therefore, the child's acceleration is 0.631 m/s².
2. The frictional force acting on the child can be calculated using the formula:
Frictional force = mass * acceleration
Mass = 49.4 kg (given)
Acceleration = 0.631 m/s² (calculated in step 1)
Frictional force = 49.4 * 0.631
= 31.11 N
Therefore, the horizontal force of friction acting on the child is 31.11 N.
3. The minimum coefficient of static friction required to prevent slipping can be calculated using the formula:
Coefficient of static friction = (tangential speed)² / (g * radius)
Tangential Speed = 1.206 m/s (calculated in step 1)
Gravity = 9.8 m/s² (given)
Radius = 2.3 m (given)
Coefficient of static friction = (1.206)² / (9.8 * 2.3)
= 0.066
Therefore, the minimum coefficient of static friction required to prevent slipping is 0.066.
To find the child's acceleration, we can use the equation of uniform circular motion. The child's acceleration is given by the centripetal acceleration, which is given by:
ac = (v^2) / r
where ac is the centripetal acceleration, v is the linear velocity, and r is the radius of the circular path.
First, let's find the linear velocity of the child. Since the merry-go-round makes one complete revolution in 12.2 seconds, we can find the angular velocity ω using the formula:
ω = (2π) / T
where T is the time for one complete revolution.
ω = (2 * π) / 12.2
Next, we can find the linear velocity v using the formula:
v = ω * r
v = ((2 * π) / 12.2) * 2.3
Once we have the linear velocity, we can now find the centripetal acceleration:
ac = v^2 / r
ac = (v^2) / 2.3
Next, let's calculate the child's acceleration:
ac = (v^2) / 2.3
Finally, we substitute the values and solve for the child's acceleration:
ac = ((2 * π / 12.2) * 2.3)^2 / 2.3
ac = ((2 * 3.1416 / 12.2) * 2.3)^2 / 2.3
ac = 0.3776^2 / 2.3
ac ≈ 0.06184 m/s^2
So, the child's acceleration is approximately 0.06184 m/s^2.
To find the horizontal force of friction acting on the child, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration:
F = m * a
where F is the force, m is the mass, and a is the acceleration.
Now, let's calculate the force of friction:
F = m * a
F = 49.4 * 0.06184
F ≈ 3.040 N
So, the horizontal force of friction acting on the child is approximately 3.040 N.
To find the minimum coefficient of static friction needed to keep the child from slipping, we need to understand the condition for an object not to slip. The maximum force of static friction can be written as:
fs ≤ μs * N
where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.
In this case, the normal force N is equal to the gravitational force acting on the child:
N = m * g
where m is the mass and g is the acceleration due to gravity.
Now, let's calculate the normal force:
N = m * g
N = 49.4 * 9.8
N ≈ 484.12 N
Now, we can find the minimum coefficient of static friction:
fs ≤ μs * N
μs * N ≥ fs
μs ≥ fs / N
Since the child is not slipping, the force of static friction fs is equal to the horizontal force of friction:
μs ≥ F / N
μs ≥ 3.040 / 484.12
μs ≈ 0.00629
So, the minimum coefficient of static friction required to keep the child from slipping is approximately 0.00629.