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PHYSICS 2231 - SOLUTION SET 1

Problem 1: Derive the acceleration ~a in polar coordinates and the unit vectors rˆ and ✓ˆ Solution to Problem 1: r d✓ˆ First, find dˆ dt and dt as they will be needed later: ⌘ ⇣ ⌘ dˆ r d ⇣ˆ = i cos ✓ + ˆj sin ✓ = ˆi✓˙ sin ✓ + ˆj ✓˙ sin ✓ = ✓˙ ˆi sin ✓ + ˆj cos ✓ = ✓˙✓ˆ dt dt ⌘ ⌘ ⇣ ˆ d✓ d ⇣ ˆ ˙r = i sin ✓ + ˆj cos ✓ = ˆi✓˙ cos ✓ ˆj ✓˙ sin ✓ = ✓˙ ˆi cos ✓ + ˆj sin ✓ = ✓ˆ dt dt Now find ~a: ✓ ◆ d~v d d~r ~a = = dt dt dt ✓ ◆ d d = (rˆ r) dt dt ✓ ◆ d dˆ r = rˆ ˙r + r dt dt ⌘ d ⇣ = rˆ ˙ r + r✓˙✓ˆ dt dˆ r d✓ˆ = r¨rˆ + r˙ + r˙ ✓˙✓ˆ + r✓¨✓ˆ + r✓˙ dt dt 2 ˙ ˙ ˆ ¨ ˆ ˙ = r¨rˆ + r˙ ✓ + r˙ ✓✓ + r✓✓ r✓ rˆ ⇣ ⌘ ⇣ ⌘ ) ~a = rˆ r¨ r✓˙2 + ✓ˆ r✓¨ + 2r˙ ✓˙ Problem 1.7: Let a ˆ and ˆb be unit vectors in the xy plane making angles ✓ and with the x axis, respectively. Show that a ˆ = cos ✓ˆi + sin ✓ˆj, ˆb = cos ˆi + sin ˆj, and using vector algebra prove that cos(✓ ) = cos ✓ cos + sin ✓ sin . 1

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PHYSICS 2231 - SOLUTION SET 1

Solution to Problem 1.7: a ˆ = aiˆi + aj ˆj ai = a ˆ · ˆi = |ˆ a||ˆi| cos ✓ = cos ✓ ⇣⇡ ⌘ aj = a ˆ · ˆj = |ˆ a||ˆj| cos ✓ = sin ✓ 2 )a ˆ = cos ✓ˆi + sin ✓ˆj ˆb = biˆi + bj ˆj bi = ˆb · ˆi = |ˆb||ˆi| cos

= cos ⌘ ⇣⇡ = sin bj = ˆb · ˆj = |ˆb||ˆj| cos 2 ) ˆb = cos ˆi + sin ˆj

) cos (✓

a ˆ · ˆb = |ˆ a||ˆb| cos (✓ ) a ˆ · ˆb = cos ✓ cos + sin ✓ sin ) = cos ✓ cos + sin ✓ sin

Problem 1.11: ~ be an arbitrary vector and let n Let A ˆ be a unit vector in some fixed direction. Show ~ ~ ~ that A = (A · n ˆ )ˆ n + (ˆ n ⇥ A) ⇥ n ˆ. Solution to Problem 1.11: Use the vector triple product identity: ~a ⇥ (~b ⇥ ~c) = ~b(~a · ~c) ~c(~a · ~b). Note that (~a ⇥ ~b) ⇥ ~c = ~c ⇥ (~a ⇥ ~b). ⇣ ⌘ ⇣ ⌘ ~ ⇥n ~ ~·n n ˆ⇥A ˆ = (ˆ n·n ˆ) A A ˆ n ˆ ⇣ ⌘ ⇣ ⌘ ~ ⇥n ~ ~·n n ˆ⇥A ˆ=A A ˆ n ˆ ~ = (A ~·n ~ ⇥n )A ˆ )ˆ n + (ˆ n ⇥ A) ˆ

Problem 1.15: By relative velocity we mean velocity with respect to a specified coordinate system. a. A point is observed to have velocity ~vA relative to coordinate system A. What is its velocity relative to coordinate system B, which is displaced from system A by distance ~ R? b. Particles a and b move in opposite directions around a circle with angular speed !, as shown. At t = 0 they are both at the point ~r = lˆj, where l is the radius of the circle. Find the velocity of a relative to b.

PHYSICS 2231 - SOLUTION SET 1

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Solution to Problem 1.15: Part a: d~rA dt d~rB ~vB = dt ~ ~rB = ~rA R(t) ~vA =

~vB =

d~rA dt

) ~vB = ~vA

~ dR dt ~ dR dt

Part b: ~ra = l sin ✓ˆi + l cos ✓ˆj ~rb = l sin ✓ˆi + l cos ✓ˆj Where ✓ is measured from the positive ˆj axis. At t = 0, ~r = lˆj. Since ✓ = !t, we can quickly check our position vectors: ~ra (t) = l sin(!t)ˆi + l cos(!t)ˆj ~ra (t = 0) = l sin(0)ˆi + l cos(0)ˆj = lˆj X ~rb (t) = l sin(!t)ˆi + l cos(!t)ˆj ~rb (t = 0) =

l sin(0)ˆi + l cos(0)ˆj = lˆj X

The relative position and velocity can be found by: ~rarelb = ~ra

~rb = 2l sin ✓ˆi

) ~rarelb = 2l sin(!t)ˆi d ~varelb = ~rarelb dt ) ~varelb = 2l! cos(!t)ˆi Problem 1.19: A tire of radius R rolls in a straight line without slipping. Its center moves with constant speed V . A small pebble lodged in the tread of the tire touches the road at t = 0. Find the pebble’s position, velocity and acceleration as functions of time.

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PHYSICS 2231 - SOLUTION SET 1

Solution to Problem 1.19: • Let ~rt be the vector from the origin (in the figure, or really any stationary point to measure the system from) to the center of the tire, • let ~rp be the vector from the origin to the pebble’s location on the tread of the tire, • and let ~rrel be the vector from the center of the tire to the pebble. ~rt = V tˆ x + Rˆ y ~rp = ~rt + ~rrel The center of the tire is moving with constant speed V , so the tire is rolling with constant angular velocity, ! = VR . Also, measure the angle !t from the ˆj axis on the center of the wheel. R sin(!t)ˆi R cos(!t)ˆj ✓ ◆ ✓ ◆ Vt ˆ Vt ˆ = sin i R cos j R R ✓ ◆ ✓ ◆ Vt ˆ Vt ˆj = V t R sin i + R R cos R R d~rp = dt ✓ ◆ ✓ ◆ Vt ˆ Vt ˆj = V V cos i + V sin R R d~vp = dt ✓ ◆ ✓ ◆ Vt ˆ Vt ˆj = V 2 sin i + V 2 cos R R

~rrel =

) ~rp ~vp ) ~vp ~ap ) ~vp

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Problem 1: Derive the acceleration ~a in polar coordinates and the unit vectors rˆ and ✓ˆ Solution to Problem 1: r d✓ˆ First, find dˆ dt and dt as they will be needed later: ⌘ ⇣ ⌘ dˆ r d ⇣ˆ = i cos ✓ + ˆj sin ✓ = ˆi✓˙ sin ✓ + ˆj ✓˙ sin ✓ = ✓˙ ˆi sin ✓ + ˆj cos ✓ = ✓˙✓ˆ dt dt ⌘ ⌘ ⇣ ˆ d✓ d ⇣ ˆ ˙r = i sin ✓ + ˆj cos ✓ = ˆi✓˙ cos ✓ ˆj ✓˙ sin ✓ = ✓˙ ˆi cos ✓ + ˆj sin ✓ = ✓ˆ dt dt Now find ~a: ✓ ◆ d~v d d~r ~a = = dt dt dt ✓ ◆ d d = (rˆ r) dt dt ✓ ◆ d dˆ r = rˆ ˙r + r dt dt ⌘ d ⇣ = rˆ ˙ r + r✓˙✓ˆ dt dˆ r d✓ˆ = r¨rˆ + r˙ + r˙ ✓˙✓ˆ + r✓¨✓ˆ + r✓˙ dt dt 2 ˙ ˙ ˆ ¨ ˆ ˙ = r¨rˆ + r˙ ✓ + r˙ ✓✓ + r✓✓ r✓ rˆ ⇣ ⌘ ⇣ ⌘ ) ~a = rˆ r¨ r✓˙2 + ✓ˆ r✓¨ + 2r˙ ✓˙ Problem 1.7: Let a ˆ and ˆb be unit vectors in the xy plane making angles ✓ and with the x axis, respectively. Show that a ˆ = cos ✓ˆi + sin ✓ˆj, ˆb = cos ˆi + sin ˆj, and using vector algebra prove that cos(✓ ) = cos ✓ cos + sin ✓ sin . 1

2

PHYSICS 2231 - SOLUTION SET 1

Solution to Problem 1.7: a ˆ = aiˆi + aj ˆj ai = a ˆ · ˆi = |ˆ a||ˆi| cos ✓ = cos ✓ ⇣⇡ ⌘ aj = a ˆ · ˆj = |ˆ a||ˆj| cos ✓ = sin ✓ 2 )a ˆ = cos ✓ˆi + sin ✓ˆj ˆb = biˆi + bj ˆj bi = ˆb · ˆi = |ˆb||ˆi| cos

= cos ⌘ ⇣⇡ = sin bj = ˆb · ˆj = |ˆb||ˆj| cos 2 ) ˆb = cos ˆi + sin ˆj

) cos (✓

a ˆ · ˆb = |ˆ a||ˆb| cos (✓ ) a ˆ · ˆb = cos ✓ cos + sin ✓ sin ) = cos ✓ cos + sin ✓ sin

Problem 1.11: ~ be an arbitrary vector and let n Let A ˆ be a unit vector in some fixed direction. Show ~ ~ ~ that A = (A · n ˆ )ˆ n + (ˆ n ⇥ A) ⇥ n ˆ. Solution to Problem 1.11: Use the vector triple product identity: ~a ⇥ (~b ⇥ ~c) = ~b(~a · ~c) ~c(~a · ~b). Note that (~a ⇥ ~b) ⇥ ~c = ~c ⇥ (~a ⇥ ~b). ⇣ ⌘ ⇣ ⌘ ~ ⇥n ~ ~·n n ˆ⇥A ˆ = (ˆ n·n ˆ) A A ˆ n ˆ ⇣ ⌘ ⇣ ⌘ ~ ⇥n ~ ~·n n ˆ⇥A ˆ=A A ˆ n ˆ ~ = (A ~·n ~ ⇥n )A ˆ )ˆ n + (ˆ n ⇥ A) ˆ

Problem 1.15: By relative velocity we mean velocity with respect to a specified coordinate system. a. A point is observed to have velocity ~vA relative to coordinate system A. What is its velocity relative to coordinate system B, which is displaced from system A by distance ~ R? b. Particles a and b move in opposite directions around a circle with angular speed !, as shown. At t = 0 they are both at the point ~r = lˆj, where l is the radius of the circle. Find the velocity of a relative to b.

PHYSICS 2231 - SOLUTION SET 1

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Solution to Problem 1.15: Part a: d~rA dt d~rB ~vB = dt ~ ~rB = ~rA R(t) ~vA =

~vB =

d~rA dt

) ~vB = ~vA

~ dR dt ~ dR dt

Part b: ~ra = l sin ✓ˆi + l cos ✓ˆj ~rb = l sin ✓ˆi + l cos ✓ˆj Where ✓ is measured from the positive ˆj axis. At t = 0, ~r = lˆj. Since ✓ = !t, we can quickly check our position vectors: ~ra (t) = l sin(!t)ˆi + l cos(!t)ˆj ~ra (t = 0) = l sin(0)ˆi + l cos(0)ˆj = lˆj X ~rb (t) = l sin(!t)ˆi + l cos(!t)ˆj ~rb (t = 0) =

l sin(0)ˆi + l cos(0)ˆj = lˆj X

The relative position and velocity can be found by: ~rarelb = ~ra

~rb = 2l sin ✓ˆi

) ~rarelb = 2l sin(!t)ˆi d ~varelb = ~rarelb dt ) ~varelb = 2l! cos(!t)ˆi Problem 1.19: A tire of radius R rolls in a straight line without slipping. Its center moves with constant speed V . A small pebble lodged in the tread of the tire touches the road at t = 0. Find the pebble’s position, velocity and acceleration as functions of time.

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PHYSICS 2231 - SOLUTION SET 1

Solution to Problem 1.19: • Let ~rt be the vector from the origin (in the figure, or really any stationary point to measure the system from) to the center of the tire, • let ~rp be the vector from the origin to the pebble’s location on the tread of the tire, • and let ~rrel be the vector from the center of the tire to the pebble. ~rt = V tˆ x + Rˆ y ~rp = ~rt + ~rrel The center of the tire is moving with constant speed V , so the tire is rolling with constant angular velocity, ! = VR . Also, measure the angle !t from the ˆj axis on the center of the wheel. R sin(!t)ˆi R cos(!t)ˆj ✓ ◆ ✓ ◆ Vt ˆ Vt ˆ = sin i R cos j R R ✓ ◆ ✓ ◆ Vt ˆ Vt ˆj = V t R sin i + R R cos R R d~rp = dt ✓ ◆ ✓ ◆ Vt ˆ Vt ˆj = V V cos i + V sin R R d~vp = dt ✓ ◆ ✓ ◆ Vt ˆ Vt ˆj = V 2 sin i + V 2 cos R R

~rrel =

) ~rp ~vp ) ~vp ~ap ) ~vp

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