Respuesta :
Answer:
a) Yes
b) 7 rad/s
c) 0.01034 J
Explanation:
a)
Yes the angular momentum of the block is conserved since the net torque on the block is zero.
b)
m = mass of the block = 0.0250 kg
w₀ = initial angular speed before puling the cord = 1.75 rad/s
r₀ = initial radius before puling the cord = 0.3 m
w = final angular speed after puling the cord = ?
r = final radius after puling the cord = 0.15 m
Using conservation of angular momentum
m r₀² w₀ = m r² w
r₀² w₀ = r² w
(0.3)² (1.75) = (0.15)² w
w = 7 rad/s
c)
Change in kinetic energy is given as
ΔKE = (0.5) (m r² w² - m r₀² w₀²)
ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)
ΔKE = 0.01034 J
a) Yes the angular momentum of the block is conserved because the net torque on the block is zero.
b) [tex]\rm \omega = 7\;rad/sec[/tex]
c) [tex]\rm \Delta KE = 0.01034\;J[/tex]
Given :
Mass, m = 0.0250 Kg
Initial radius before puling the cord, [tex]\rm r_0[/tex] = 0.3 m
Angular speed before pulling the cord, [tex]\rm \omega_0[/tex] = 1.75 rad/sec
Final radius after puling the cord, r = 0.15 m
Solution :
a) Yes the angular momentum of the block is conserved because the net torque on the block is zero.
b) To find the new angular speed, conserve the angular momentum,
[tex]\rm mr_0^2\omega_0 = m r^2\omega[/tex] --- (1)
where, [tex]\omega[/tex] is the final angular speed after puling the cord.
Now put the values of [tex]\rm \omega_0, \; r_0, \; r \; and \; m[/tex] in equation (1) we get,
[tex](0.3)^2\times 1.75 = (0.15)^2\times \omega[/tex]
[tex]\omega = \dfrac{0.3^2 \times 1.75 }{0.15^2}[/tex]
[tex]\rm \omega = 7\;rad/sec[/tex]
c) We know that change in kinetic energy is given by,
[tex]\rm \Delta KE = \dfrac{1}{2}mr^2\omega^2 - \dfrac{1}{2}mr_0^2\omega_0^2[/tex]
[tex]\rm \Delta KE = \dfrac{1}{2}\times 0.025\times((0.15^2\times 7^2)- (0.3^2 \times 1.75^2))[/tex]
[tex]\rm \Delta KE = 0.01034\;J[/tex]
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https://brainly.com/question/17858145?referrer=searchResults
