Answer :
Answer:
a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J
Explanation:
a) If the student completes one turn in 1.26 sec, this is called the period of the movement.
If we take into account that the angle rotated during one turn, is 2π rads, by definition of angular velocity, we can get this value as follows:
ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.
b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:
Li = Lf ⇒ I₁ * ω₁ = I₂* ω₂
So, we can solve for ω₂, as follows:
ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec
c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:
W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²
W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)
W = 7.03 J
Initial angular speed of the system is 4.987 radian/sec
New angular speed of the system is 5.576
What is angular speed?
Angular speed is a term used to describe how a circular objects spins
Given data
time, t = 1.26 sec
moment of Inertia, I = 2.25 kgm^2
distance covered, θ = 360° = 2π radians
Angular speed, ω = ?
a) calculating initial Angular speed, ω
ω = θ / t
ω = 2π / 1.26
ω = `4.987 radians/sec
b) calculating new angular speed, ω2
Energy generated during the circular motion, k
k = (1 / 2) I ω^2
k = 0.5 * 2.25 * 4.987^2 = 27.979 J
assuming energy was conserved, the new moment of inetia and angular speed is substituted into the the energy hence
k = 27.979 = (1 / 2) I2 ω2^2
27.979 = 0.5 * 1.80 * ω2^2
ω2^2 = 27.979 / (0.5 * 1.8)
ω2 = sqrt(31.088)
ω2 = 5.576 radians / sec
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