1. A science teacher wanting to demonstrate centripetal acceleration holds a bucketful of water
by its handle and spins it in a vertical circle. If the teacher’s arm is 1.15m long, at what speed
must the bucket be rotated so that at the top of its path there is no tension acting through the
teacher’s arm? (2 marks)
2. An amusement park ride spins children on swings in a horizontal circle with
a radius of 6.2 m (see diagram in the top right corner of the activity). If a 21 kg
child completes 11 rotations in 2 minutes
a. identify the centripetal force acting on the child; (1 mark)
b. determine the speed of the child; and (2 marks)
c. determine the tension in the swing chain. (2 marks)
3. You are designing a model roller coaster for a school project with a loop of
radius 62 cm. What is the minimum cart speed so that the cart does not fall off
the track at the top of the loop? (4 marks)
4. A 1200 kg Indy car is travelling around a curve with a radius of 95 m. If the
curve is banked at 24° and the coefficient of friction is 0.28, what is the
maximum speed at which the car can travel without flying off?
1. v^2/r = g
2. omega in radians/second
= 2 pi * 11/120
a.
Ac = r omega^2
F = m Ac = m r omega^2
b.
v = omega * r
c.
same as a.
3. r = .62 meters
same as problem 1.
4. Normal force = m g cos 24
max friction force = .28 * m g cos 24
component of mass down slope = m g sin 24
so force holding car from slipping up
= m g (.28 cos 24 + sin 24)
that has to equal m v^2/R cos 24
jnm
1. To find the speed at which the bucket must be rotated so that there is no tension acting through the teacher's arm at the top of its path, we can use the concept of centripetal acceleration.
Centripetal acceleration is given by the formula: ac = (v^2) / r, where ac is the centripetal acceleration, v is the velocity, and r is the radius of the circular path.
At the top of the path, the tension force is zero, so the centripetal force is equal to the weight of the water in the bucket. The centripetal force is given by the formula: Fc = m * ac, where Fc is the centripetal force, m is the mass of the water, and ac is the centripetal acceleration.
Since the weight of the water is equal to the mass of the water multiplied by the acceleration due to gravity (9.8 m/s^2), we can equate the two forces:
Fc = m * ac = m * (v^2) / r
Rearranging the equation, we can solve for v:
v^2 = Fc * r / m
v = sqrt( Fc * r / m )
Substituting the known values (r = 1.15 m and Fc = m * g = m * 9.8 m/s^2), we can calculate the speed at which the bucket must be rotated.
2.
a. The centripetal force acting on the child is provided by the tension in the swing chain.
b. The speed of the child can be calculated using the formula for centripetal acceleration:
ac = (v^2) / r
Solving for v:
v = sqrt(ac * r)
The child completes 11 rotations in 2 minutes. To find the angular speed (ω) of the child, we can use the formula:
ω = (2π * n) / t
where n is the number of rotations and t is the time taken.
ω = (2π * 11) / (2 * 60)
Now we can calculate the centripetal acceleration (ac) using the formula:
ac = ω^2 * r
Plugging in the values (r = 6.2 m and ω calculated above), we can find the speed of the child.
c. The tension in the swing chain can be calculated using the centripetal force formula:
Fc = m * ac
Where m is the mass of the child and ac is the centripetal acceleration calculated in part b.
3. To calculate the minimum cart speed so that the cart does not fall off the track at the top of the loop, we need to consider the forces acting on the cart. At the top of the loop, the centripetal force is provided by the normal force.
The centripetal force is given by the formula: Fc = m * ac, where Fc is the centripetal force, m is the mass of the cart, and ac is the centripetal acceleration.
Since the centripetal force is provided by the normal force, we can equate the two forces:
Fc = m * ac = m * (v^2) / r
Solving for v:
v^2 = ac * r
v = sqrt(ac * r)
The minimum cart speed is the speed at which the centripetal force required to stay on the track is equal to the gravitational force pulling down on the cart (m * g).
So, ac * r = m * g
Substituting the known values (r = 62 cm = 0.62 m and g = 9.8 m/s^2), we can calculate the minimum cart speed.
4. To find the maximum speed at which the car can travel without flying off the curve, we need to consider the forces acting on the car. The centripetal force required to keep the car moving along the curved path is provided by the friction force between the tires and the road.
The centripetal force is given by the formula: Fc = m * ac, where Fc is the centripetal force, m is the mass of the car, and ac is the centripetal acceleration.
The friction force can be calculated using the formula: Ff = μ * N, where Ff is the friction force, μ is the coefficient of friction, and N is the normal force.
At the maximum speed (v), the friction force is equal to the maximum possible friction force. The maximum possible friction force is given by: Ff_max = μ * m * g
To calculate the centripetal force at the maximum speed, we equate the two forces:
Fc = Ff_max
m * ac = μ * m * g
Solving for ac:
ac = μ * g
The centripetal acceleration at the maximum speed is equal to μ * g.
Now we can calculate the maximum speed (v) using the formula for centripetal acceleration:
ac = (v^2) / r
v = sqrt(ac * r)
Substituting the known values (r = 95 m and ac = μ * g), we can calculate the maximum speed at which the car can travel without flying off the curve.