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Answer:
(a) the angular velocity at θ1 is 11.64 rad/s
(b) the angular acceleration is 0.12 rad/
(c) the angular position was the disk initially at rest is - 428.27 rad
Explanation:
Given information :
θ1 = 16 rad
θ2 = 76 rad
ω2 = 11 rad/s
t = 5.3 s
(a) The angular velocity at θ1
First, we use the angular motion equation for constant acceleration
Δθ = (ω1+ω2)t/2
θ2 - θ1 = (ω1+ω2)t/2
ω1 + ω2 = 2 (θ2 - θ1) / t
ω1 = (2 (θ2 - θ1) / t ) - ω2
= (2 (76-16) / 5.3) - 11
= 11.64 rad/s
(b) the angular acceleration
ω2 = ω1 + α t
α t = ω2 - ω1
α = (ω2 - ω1)/t
= (11.64 - 11) / 5.3
= 0.12 rad/
(c) the angular position was the disk initially at rest, θ0
at rest ω0 = 0
ω2^2 = ω01 t + 2 α Δθ
2 α Δθ = ω2^2
θ2 - θ0 = ω2^2 / 2 α
θ0 = θ2 - (ω2^2) / 2 α
= 76 - (/ 2 x 0.12
= 76 - 504.16
= - 428.27 rad
Answer:
a. 0.21 rad/s2
b. 2.205 N
Explanation:
We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds
200 rpm = 200 * 2π / 60 = 21 rad/s
180 rpm = 180 * 2π / 60 = 18.85 rad/s
r = d/2 = 30cm / 2 = 15 cm = 0.15 m
a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is
b) Assume the grind stone is a solid disk, its moment of inertia is
Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.
So the friction torque is
The friction force is
Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone