A turntable reaches an angular speed of 33.3 rpm, in 2.0 s, starting from rest.
(a) Assuming the angular acceleration is constant, what is its magnitude?
(b) How many revolutions does the turntable make during this time interval?
a.
Angular acceleration
=(ωfinal-ωinitial)/time
=(33.3*2π radians/60 sec. - 0)/2 sec.
= 1.74 rad/s/s
b.
Number of turns
= (ωfinal+ωinitial)/2 * time radians
note: ω is in rad/s
To solve this problem, we can use the equations of rotational motion.
(a) To find the magnitude of the angular acceleration, we can use the following equation:
ω = ω_0 + αt
where:
ω is the final angular velocity (in radians per second),
ω_0 is the initial angular velocity (in radians per second),
α is the angular acceleration (in radians per second squared), and
t is the time (in seconds).
Given:
ω = 33.3 rpm
ω_0 = 0 (starting from rest)
t = 2.0 s
Step 1: Convert 33.3 rpm to radians per second
1 revolution = 2π radians
33.3 revolutions/minute * (2π radians / 1 revolution) * (1 minute / 60 seconds) = 33.3 * 2π / 60 radians/second ≈ 3.49 radians/second
Step 2: Substitute the known values into the equation to solve for α
3.49 = 0 + α(2.0)
2α = 3.49
α ≈ 1.745 radians/second squared
Therefore, the magnitude of the angular acceleration is approximately 1.745 radians/second squared.
(b) To find the number of revolutions the turntable makes during this time interval, we can use the following equation:
θ = ω_0t + (1/2)αt^2
where:
θ is the angular displacement (in radians),
ω_0 is the initial angular velocity (in radians per second),
α is the angular acceleration (in radians per second squared), and
t is the time (in seconds).
Given that ω_0 = 0 (starting from rest) and t = 2.0 s, we can simplify the equation to:
θ = (1/2)αt^2
Substituting the known values, we have:
θ = (1/2)(1.745)(2.0)^2
θ = (1/2)(1.745)(4.0)
θ ≈ 3.49 radians
To convert this to revolutions, we can use the conversion factor:
1 revolution = 2π radians
θ ≈ 3.49 radians * (1 revolution / 2π radians) ≈ 0.555 revolutions
Therefore, the turntable makes approximately 0.555 revolutions during this time interval.
To solve this problem, we can use the equations of rotational motion. The relevant equations are:
1. Angular speed (ω) = initial angular speed (ω₀) + angular acceleration (α) × time (t)
2. Angular displacement (θ) = initial angular speed (ω₀) × time (t) + 0.5 × angular acceleration (α) × time²
(a) To find the magnitude of the angular acceleration (α), we can use the first equation. We are given the initial angular speed (ω₀) as zero, the final angular speed (ω) as 33.3 rpm, and the time (t) as 2.0 seconds. We need to convert the angular speed from rpm to radians per second. Since 1 rpm = 2π radians per minute, we have:
Initial angular speed (ω₀) = 0 radians/second
Final angular speed (ω) = 33.3 rpm × 2π radians/minute × 1/60 minutes/second = 3.49 radians/second
Time (t) = 2.0 seconds
Using the first equation, we can solve for the angular acceleration (α):
ω = ω₀ + αt
3.49 radians/second = 0 radians/second + α × 2.0 seconds
α = 3.49 radians/second / 2.0 seconds
α = 1.745 radians/second²
Therefore, the magnitude of the angular acceleration is 1.745 radians/second².
(b) To find the number of revolutions the turntable makes during the time interval, we need to calculate the angular displacement (θ). We can use the second equation, but since the initial angular speed (ω₀) is zero, the equation simplifies to:
θ = 0.5 × α × time²
Using the given values, we have:
θ = 0.5 × 1.745 radians/second² × (2.0 seconds)²
θ = 1.745 radians/second² × 4 seconds²
θ = 6.98 radians
To convert the angular displacement to revolutions, we divide by 2π radians per revolution:
Number of revolutions = 6.98 radians / (2π radians/revolution)
Number of revolutions ≈ 1.11 revolutions
Therefore, the turntable makes approximately 1.11 revolutions during the 2.0-second time interval.