A bob attached to a string of length 2 m, is displaced by an angle of 8° and then released
from rest. The maximum angular acceleration, θ"max, is: (Take g = 10 m/s2)
Mark only one oval.
0.56 rad/s2
0.7 rad/s2
0.42 rad/s2
0.8 rad/s2
0.25 rad/s2
To find the maximum angular acceleration, we need to use the formula for angular acceleration. Angular acceleration is defined as the rate of change of angular velocity with respect to time.
The formula for angular acceleration is:
θ" = ω^2 / r
Where:
θ" represents the angular acceleration
ω represents the angular velocity
r represents the radius or length of the string
In this question, the maximum angular acceleration, θ"max, is what we need to find. We are given the length of the string, which is 2 m, and the angular displacement, which is 8° or 8/360 radians.
First, let's convert the given angle to radians:
θ = 8° = (8/360) radians
Now, we need to find the angular velocity, ω. Since the bob is released from rest, the initial angular velocity is 0. We can find the final angular velocity using the equation:
ω^2 = 2 * θ * θ"
0 = 2 * (8/360) * θ"
0 = (16/360) * θ"
Now, let's solve for θ"max:
θ"max = -360 / 16 radians/second^2 (We use a negative sign since the direction is opposite to the displacement)
Let's simplify this value:
θ"max = -22.5 radians/second^2
Since the options provided are in rad/s^2 instead of radians/second^2, we need to convert the value:
θ"max = -22.5 rad/second^2 * (1/10) (Since 1 radian = 10 rad/second^2)
θ"max = -2.25 rad/second^2
Since θ"max represents the maximum angular acceleration, it can't be negative. Therefore, we take the absolute value:
θ"max = 2.25 rad/second^2
From the given options, the closest value to 2.25 rad/s^2 is 0.56 rad/s^2.
Therefore, the correct oval to mark is:
0.56 rad/s^2.