An EF 151 GTA sits on the EF stool and rotates CCW with an initial speed of 60 rpm. After 28 revolutions he stops. Assuming constant acceleration, how long was he rotating?
Before there was central heating, hot water bottles were used to keep people warm at night. These flat containers were filled with hot water and placed at the foot of the bed, under the blankets. How was the heat energy transferred from the hot water bottle into the space under the blankets?
convection
W2DFW
To determine how long the EF 151 GTA was rotating, we need to calculate the time taken for 28 revolutions with a constant acceleration.
To solve this problem, we can break it down into three steps:
Step 1: Convert the initial speed from rpm to rad/s.
Step 2: Calculate the angular acceleration.
Step 3: Use the kinematic equation to find the time.
Let's go through each step in detail:
Step 1: Convert the initial speed from rpm to rad/s.
The initial speed is given as 60 rpm (revolutions per minute). To convert this to radians per second, multiply by 2π/60:
Initial speed = (60 rpm) x (2π rad/1 min) / (60 min/1 hour) x (1 hour/3600 sec)
= 60 x 2π / 3600
= π / 30 rad/s
Step 2: Calculate the angular acceleration.
In this problem, we assume constant acceleration. Since we are given the initial and final speeds and rotations, we can use the following equation to calculate the angular acceleration:
ωf^2 = ωi^2 + 2αθ
Where ωi is the initial angular velocity (converted to rad/s), ωf is the final angular velocity (0 rad/s when he stops), α is the angular acceleration, and θ is the number of rotations (28 revolutions).
Substituting the known values:
0^2 = (π/30)^2 + 2α(28)
0 = (π^2 / 900) + 56α
Rearranging the equation:
56α = -π^2 / 900
α = -π^2 / (900 * 56) rad/s²
Step 3: Use the kinematic equation to find the time.
The kinematic equation for rotational motion is:
θ = ωi * t + 0.5 * α * t^2
Since the final angular position (θ) is given as 28 revolutions, we can convert this to radians:
28 revolutions * (2π rad/1 revolution) = 56π rad
Now, we can rearrange the equation to solve for time (t):
56π = (π/30) * t + 0.5 * (-π^2 / (900 * 56)) * t^2
Simplifying further:
56π = (π/30) * t - (π^2 / (900 * 56)) * (t^2)
Multiplying through by 30 * (900 * 56):
(900 * 56) * 56π = 900 * 56π * t - π^2 * t^2
Simplifying:
(900 * 56 * 56)π = (900 * 56) * π * t - π^2 * t^2
Dividing both sides by π:
900 * 56 * 56 = 900 * 56 * t - π * t^2
Dividing both sides by 900 * 56:
t^2 - t * (900 * 56) + (900 * 56 * 56) = 0
This is a quadratic equation that can be solved using the quadratic formula, but the math becomes cumbersome. However, we can simplify the problem further by noting that the quadratic equation has a perfect square trinomial form:
t^2 - t * (900 * 56) + (900 * 56 * 56) = (t - (900 * 28))^2
Since the square of a number is always positive, we can conclude that t - (900 * 28) = 0, when solving for the positive value of t:
t = 900 * 28 = 25200
Therefore, the time he was rotating for is 25200 seconds or 7 hours.