Harry Potter decides to take Pottery 101 as an elective to satisfy his arts requirement at Hogwarts. He sets some clay
(m = 4.00 kg)
on the edge of a pottery wheel
(r = 0.730 m),
which is initially motionless. He then begins to rotate the wheel with a uniform acceleration, reaching a final angular speed of 2.800 rev/s in 2.80 s.
(a) What is the speed of the clay when the initial 2.80 s has passed?
m/s
(b) What is the centripetal acceleration of the clay initially and when the initial 2.80 s has passed? (Enter the magnitudes of your answers.)
ac,i = m/s^2
ac,f = m/s^2
(c) What is the magnitude of the constant tangential acceleration responsible for starting the clay in circular motion?
m/s^2
I am reluctant to do these without seeing your work.
a. SPEED=w*r=2PI*2.800*radius
b. initially? isn't it zero? the wheel is at stop.
final? w^2*r=(2PI*2.8)^2* radius
c. tangential acceleration=changeradialvelociy/time
= 2PI*2.8*radius/time
Well, Harry Potter sure knows how to mix magic with art! Let's get to your questions:
(a) What is the speed of the clay when the initial 2.80 s has passed?
To find the speed, we need to use the formula:
speed = (angular speed) * (radius)
So, the speed of the clay is:
speed = (2.800 rev/s) * (0.730 m) = 2.044 m/s
(b) What is the centripetal acceleration of the clay initially and when the initial 2.80 s has passed?
The centripetal acceleration can be calculated using the formula:
centripetal acceleration = (angular speed)^2 * (radius)
Initially, when the pottery wheel is motionless, the centripetal acceleration is zero:
ac,i = 0 m/s^2
After 2.80 s, when the wheel has reached its final angular speed, the centripetal acceleration is:
ac,f = (2.800 rev/s)^2 * (0.730 m) = 6.989 m/s^2
(c) What is the magnitude of the constant tangential acceleration responsible for starting the clay in circular motion?
To find the tangential acceleration, we can use the formula:
tangential acceleration = (change in angular speed) / (time)
Since the clay starts from rest and reaches a final angular speed of 2.800 rev/s in 2.80 s, the change in angular speed is:
(change in angular speed) = (2.800 rev/s)
Using the formula, the magnitude of the tangential acceleration is:
tangential acceleration = (2.800 rev/s) / (2.80 s) = 1 rev/s^2
However, just a friendly note: In the magical world of Harry Potter, they often use magic to make things happen, not just physics formulas! So, don't be surprised if Harry finds a way to create pottery in a snap with a flick of his wand!
To solve this problem, we can use the formulas for angular motion and circular motion.
(a) To find the speed of the clay after 2.80 seconds, we can use the formula:
ω = ω₀ + αt
where ω is the final angular speed, ω₀ is the initial angular speed, α is the angular acceleration, and t is the time.
Let's plug in the given values:
- ω = 2.800 rev/s (convert to rad/s by multiplying by 2π)
- α = (ω - ω₀) / t = (2.800 rev/s - 0 rev/s) / 2.80 s = 1 rev/s^2 (convert to rad/s^2 by multiplying by 2π)
Now we can rearrange the formula to solve for ω₀:
ω₀ = ω - αt
ω₀ = 2.800 rev/s - (1 rev/s^2 * 2.80 s) = 2.800 rev/s - 2.800 rev/s = 0 rev/s
The initial angular speed is 0 rev/s, which means the clay is initially stationary, so its speed is 0 m/s.
Therefore, the speed of the clay when the initial 2.80 s has passed is 0 m/s.
(b) The centripetal acceleration (ac) is given by the formula:
ac = ω²r
where ω is the angular speed and r is the radius.
Initially, when the clay is stationary, the angular speed is 0 rev/s, so ac,i = 0 m/s^2.
After 2.80 seconds, the angular speed is 2.800 rev/s (convert to rad/s):
ac,f = (2.800 rev/s)^2 * 0.730 m = 6.178 m/s^2
Therefore, the centripetal acceleration initially is 0 m/s^2, and after 2.80 seconds, it is 6.178 m/s^2.
(c) The constant tangential acceleration responsible for starting the clay in circular motion is given by the formula:
at = αr
where α is the angular acceleration and r is the radius.
Given that α = 1 rev/s^2 (convert to rad/s^2) and r = 0.730 m, we can calculate:
at = 1 rev/s^2 * 0.730 m = 0.730 m/s^2
Therefore, the magnitude of the constant tangential acceleration responsible for starting the clay in circular motion is 0.730 m/s^2.
To answer these questions, we need to use the equations relating rotational motion and circular motion:
1. For the speed of the clay:
The final angular speed of the pottery wheel is given as 2.800 rev/s. Since one revolution (rev) is equivalent to 2π radians, we can convert it to angular speed in radians per second:
ω = 2.800 rev/s * 2π rad/rev = 17.60 rad/s
The initial angular speed of the pottery wheel is 0 rad/s since it starts from rest. The angular acceleration can be found using the formula:
α = (ω - ω0) / t = (17.60 rad/s - 0 rad/s) / 2.80 s = 6.286 rad/s^2
To find the speed of the clay when the initial 2.80 s has passed, we can use the formula for linear velocity in terms of angular velocity:
v = ω * r
where "r" is the radius of the pottery wheel. Substituting in the values:
v = 17.60 rad/s * 0.730 m = 12.808 m/s
Therefore, the speed of the clay after the initial 2.80 s has passed is 12.808 m/s.
2. For the centripetal acceleration:
The centripetal acceleration (ac) can be calculated using the formula:
ac = r * α
(a) Initially:
ac,i = 0.730 m * 6.286 rad/s^2 = 4.589 m/s^2
(b) After the initial 2.80 s:
ac,f = 0.730 m * 6.286 rad/s^2 = 4.589 m/s^2
Therefore, the centripetal acceleration of the clay is 4.589 m/s^2, both initially and after 2.80 s.
3. For the magnitude of the constant tangential acceleration:
The tangential acceleration (at) can be calculated using the formula:
at = r * α
Thus, the magnitude of the constant tangential acceleration responsible for starting the clay in circular motion is equal to the centripetal acceleration, which is 4.589 m/s^2.