Question

Notify Me Bookmark

At a particular instant, a 1.0 kg particle's position is r→ = (4.0 i^−5.0 j^+2.0 k^)m, its velocity is v→ = (−1.0 i^+5.0 j^+1.0 k^)m/s, and the force on it is F→ = (13.0 i^+17.0 j^)N. (Express your answers in vector form.) (a) What is the angular momentum (in kg⋅m2 /s) of the particle about the origin? ı→ = kg⋅m2 /s (b) What is the torque (in N⋅m) on the particle about the origin? τ→ = N⋅m (c) What is the time rate of change of the particle's angular momentum about the origin at this instant (in kg⋅m2 /s2)? dl→dt = kg⋅m2 /s2

At a particular instant, a 1.0 kg particle's position is r→ = (4.0 i^−5.0 j^+2.0 k^)m, its velocity is v→ = (−1.0 i^+5.0 j^+1.0 k^)m/s, and the force on it is F→ = (13.0 i^+17.0 j^)N. (Express your answers in vector form.) (a) What is the angular momentum (in kg⋅m2 /s) of the particle about the origin? ı→ = kg⋅m2 /s (b) What is the torque (in N⋅m) on the particle about the origin? τ→ = N⋅m (c) What is the time rate of change of the particle's angular momentum about the origin at this instant (in kg⋅m2 /s2)? dl→dt = kg⋅m2 /s2

Image text
At a particular instant, a 1.0 kg particle's position is r → = ( 4.0 i ^ − 5.0 j ^ + 2.0 k ^ ) m , its velocity is v → = ( − 1.0 i ^ + 5.0 j ^ + 1.0 k ^ ) m / s , and the force on it is F → = ( 13.0 i ^ + 17.0 j ^ ) N . (Express your answers in vector form.) (a) What is the angular momentum (in k g ⋅ m 2 / s ) of the particle about the origin?
ı → = ◻ k g ⋅ m 2 / s
(b) What is the torque (in N ⋅ m ) on the particle about the origin?
τ → = ◻ N ⋅ m
(c) What is the time rate of change of the particle's angular momentum about the origin at this instant (in k g ⋅ m 2 / s 2 )?
d l → d t = ◻ k g ⋅ m 2 / s 2

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

A small space probe of mass 200 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.4 seconds after it is launched, the probe is at location < 4400, 8200, 0 > m, and at this same instant its momentum is < 48000, −8500, 0 > kg⋅m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is < −4350, −730, 0 > N. Part 1 (a) Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.4 seconds after the probe is launched to 22.8 seconds after the launch? Δp→ = < > kg⋅m/s Save for Later Attempts: 0 of 10 used Submit Answer Part 2 (b) What is the momentum of the probe at time 22.8 seconds after launch? i i kg⋅m/s Save for Later Attempts: 0 of 10 used Submit Answer Part 3 (c) Assuming that the force is nearly constant during the time interval from 22.4 seconds after the probe is launched to 22.8 seconds after the launch, so that the velocity is changing at a constant rate, what is the change of the position of the probe during this time interval? Divide the time interval into two equal parts and use the approximation v→avg ≈ p→f/m. Δr→ = < > m Save for Later Attempts: 0 of 10 used Submit Answer Part 4 (d) What is the location of the probe 22.8 seconds after launch? r→f = < iii > m eTextbook and Media Save for Later Attempts: 0 of 10 used Submit Answer

A small space probe of mass 200 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.4 seconds after it is launched, the probe is at location < 4400, 8200, 0 > m, and at this same instant its momentum is < 48000, −8500, 0 > kg⋅m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is < −4350, −730, 0 > N. Part 1 (a) Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.4 seconds after the probe is launched to 22.8 seconds after the launch? Δp→ = < > kg⋅m/s Save for Later Attempts: 0 of 10 used Submit Answer Part 2 (b) What is the momentum of the probe at time 22.8 seconds after launch? i i kg⋅m/s Save for Later Attempts: 0 of 10 used Submit Answer Part 3 (c) Assuming that the force is nearly constant during the time interval from 22.4 seconds after the probe is launched to 22.8 seconds after the launch, so that the velocity is changing at a constant rate, what is the change of the position of the probe during this time interval? Divide the time interval into two equal parts and use the approximation v→avg ≈ p→f/m. Δr→ = < > m Save for Later Attempts: 0 of 10 used Submit Answer Part 4 (d) What is the location of the probe 22.8 seconds after launch? r→f = < iii > m eTextbook and Media Save for Later Attempts: 0 of 10 used Submit Answer
Solution
0 0

Two small objects each of mass m = 0.6 kg are connected by a lightweight rod of length d = 0.7 m (see the figure). At a particular instant they have velocities whose magnitudes are v1 = 37 m/s and v2 = 57 m/s and are subjected to external forces whose magnitudes are F1 = 73 N and F2 = 32 N. The distance h = 0.2 m, and the distance w = 0.8 m. The system is moving in outer space. (a) What is the total (linear) momentum P→total of this system? P→total = kg⋅m/s (b) What is the velocity v→cm of the center of mass? v→cm = m/s (c) What is the total angular momentum L→A of the system relative to point A? L→A = kg⋅m2 /s (d) What is the rotational angular momentum L→rot of the system? L→rot = kg⋅m2 /s (e) What is the translational angular momentum L→trans of the system relative to point A? L→trans = kg⋅m2 /s (f) After a short time interval Δt = 0.23 s, what is the total (linear) momentum P→ total of the system? P→total = kg⋅m/s
Solution
0 0

A jetliner of mass 20,000 kg is flying over level ground at a constant altitude of 4.30 km with a constant velocity of 155 m/s west. Its direction is currently straight toward the top of a mountain. At a particular instant, a house is directly below the plane at ground level. (a) At this instant, what is the magnitude of the plane's angular momentum relative to the house (in kg⋅m2/s )? kg⋅m2 /s (b) As the plane continues its constant velocity motion, how does the angular momentum vector relative to the house change? The magnitude of the angular momentum vector stays the same, but the direction changes. Both the magnitude and direction of the angular momentum vector change. The direction of the angular momentum vector stays the same, but the magnitude changes. Both the magnitude and direction of the angular momentum vector stay the same. (c) At this same instant, what is the magnitude of the plane's angular momentum relative to the mountaintop (in kg⋅m2 /s )? kg⋅m2/s

Solution
0 0