Wednesday, May 9, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Forces - Problems





Notes to a video lecture on http://www.unizor.com



Forces - Problems



Problem 1



A car of mass m uniformly accelerates from a state of rest to a maximum speed vmax along a straight road during time tmax,

(a) What is the most convenient reference frame to solve this problem?

(b) What is the vector of acceleration a(t) as a function of time?

(c) What is the force F(t) its engine produces to push car forward as a function of time?

(d) What is the vector of velocity v(t) as a function of time?

(e) What is the distance d(t) it covers as a function of time?



Answers:



(a) Reference frame:

Origin at the beginning position of a car on a road, X-axis goes along
the road in the direction of car's movement, Y- and Z-axes are in any
position perpendicular to X-axis.

In this frame the Y- and Z-coordinates will always be zero for any moment of time t.

(b) Acceleration: vector directed along the X-axis with magnitude

x"(t) = a = |a(t)| = vmax /tmax

(c) Force: vector directed along the X-axis with magnitude

|F(t)| = m·a = m·vmax /tmax

(d) Velocity: vector directed along the X-axis with magnitude

x'(t) = |v(t)| = a·t = vmax ·t /tmax

(e) Distance:

x(t) = d(t) = a·t²/2 = vmax ·t²/(2·tmax)



Checking:



Acceleration must be constant:

x"(t) = vmax /tmax

(does not depend on t)

At t=0 distance from the beginning position must be zero:

x(0) = vmax ·0²/(2tmax) = 0

At t=0 vector of velocity must have magnitude of zero:

x'(0) = vmax ·0/tmax = 0

At t=tmax vector of velocity must have magnitude vmax:

x'(tmax) = vmax ·tmax/tmax = vmax





Problem 2



Exactly the same as Problem 1 above, except in the beginning at time t=0 a car is not at rest, but moves with speed vmin.



Answers:



(a) Reference frame:

Origin at the beginning position of a car on a road, X-axis goes along
the road in the direction of car's movement, Y- and Z-axes are in any
position perpendicular to X-axis.

In this frame the Y- and Z-coordinates will always be zero for any moment of time t.

(b) Acceleration: vector directed along the X-axis with magnitude

x"(t) = a = |a(t)| = (vmax−vmin)/tmax

(c) Force: vector directed along the X-axis with magnitude

|F(t)| = m·a = m·(vmax−vmin)/tmax

(d) Velocity: vector directed along the X-axis with magnitude

x'(t) = |v(t)| = vmin + a·t = vmin + (vmax−vmin)·t /tmax

(e) Distance:

x(t) = d(t) = vmin·t + a·t²/2 = vmin·t + (vmax−vmin)·t²/(2·tmax)



Checking:



Acceleration must be constant:

x"(t) = (vmax−vmin)/tmax

(does not depend on t)

At t=0 distance from the beginning position must be zero:

x(0) = vmin·0 + (vmax−vmin)·0²/(2·tmax) = 0

At t=0 vector of velocity must have magnitude of vmin:

x'(0) = vmin + (vmax−vmin)·0 /tmax = vmin

At t=tmax vector of velocity must have magnitude vmax:

x'(tmax) = vmin + (vmax−vmin)·tmax /tmax = vmax





Problem 3



A planet of mass m uniformly (that is, with constant angular speed) rotates around its star on a constant distance R from it, making one full circle (that is, radians) in to time.

(a) What is the most convenient reference frame to solve this problem?

(b) What is the angular speed ω of a planet as it rotates around its star?

(c) What is the position vector of a planet as it goes along its
circular trajectory in XY-coordinates (Z-coordinate is always 0 for
properly chosen frame of reference)?

(d) What is the vector of velocity of a planet as it goes along its
circular trajectory, and how is this vector directed relative to
position vector?

(e) What is the magnitude of velocity vector (speed) v?

(f) What is the vector of acceleration of a planet as it goes along its
circular trajectory, and how is this vector directed relative to
position vector?

(g) What is the magnitude of acceleration vector a?

(h) Express the magnitude of acceleration vector a in terms of radius of a circular trajectory R and the magnitude of the velocity vector v.

(i) What is the vector of force of gravity from a star to a planet in
terms of mass, radius of trajectory and speed (angular and linear)?



Answers:



(a) Reference frame:

Origin at the center of a star, XY-plane should be a plane where a
planet rotates, Z-axis is perpendicular to XY-plane, X-axis is from a
star to a point where planet is located at moment of time t=0.

In this frame the Z-coordinate will always be zero for any moment of time t.

(b) Angular speed:

ω = 2π/to - constant

(c) Position vector {x(t), y(t)}:

x(t) = R·cos(ωt)

y(t) = R·sin(ωt)

(d) Velocity {x'(t), y'(t)}

(first derivative of position):

x'(t) = −R·ω·sin(ωt)

y'(t) = R·ω·cos(ωt)

This velocity vector is perpendicular to position vector since their scalar product x(t)·x'(t)+y(t)·y'(t) equals to zero.

(e) Magnitude of velocity vector (speed):

v = √x'(t)²+y'(t)² = R·ω

(f) Acceleration {x"(t), y"(t)}

(second derivative of position or

first derivative of velocity):

x"(t) = −R·ω²·cos(ωt)

y"(t) = −R·ω²·sin(ωt)

This velocity vector is collinear to position vector, but directed
towards the origin of coordinates because both components have negative
multiplier −ω² in front of them.

(g) Magnitude of acceleration vector:

a = √x"(t)²+y"(t)² = R·ω²

(h) a = v²/R

(i) F = m·a = m·R·ω² = m·v²/R

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