Study GuideChapter 5

Sections 3 and 4

If the speed changes while an object is traveling in a circle it has two types of acceleration:

centripetal - due to direction change – toward center of the circletangential – due to speed change – tangent to the circle

total accel = vector sum of aT + aC

total force = vector sum of FT + Fc

A small sphere of mass m is attached to the end of a cord of length R, which rotates under the influence of the gravitational force in a vertical circle about a fixed point O. Let us determine the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle Θ with the vertical.

Case 1R α v approx true for objects that fall through

the liquid at low speed or very smallobjects in airex – sphere in honey, dust in air

R = bvSo if ΣF = ma mg – R = ma mg – bv = m(dv/dt)

dv/dt = g – (b/m)v

Case 1 cont.

dv/dt = g – (b/m)v

Note: when a = dv/dt = 0 g = b/mv mg/b = v

v is the terminal speed object no longer accelerates

since R = W

Case 1 cont.

A solution for our differential equation isv = mg/b (1 – e-bt/m)

Since vT = mg/b

v = vT (1 – e-t/Τ) and T = m/b- time constant

Time constant – time it takes for 1 – e-t/T to become equal to 1 – e-1 = 0.632

or the time for v = 63.2%vT

A small sphere of mass 2 g is released from rest in a large vessel filled with oil. The sphere approaches a terminal speed of 5 cm/s. Determine (a) the time constant and (b) the time it takes the sphere to reach 90% of its terminal speed.

Case 2

R α v2 true for large objects at high speedex – airplane, skydiver, baseball

R = ½ DρAv2 ρ = density of airA = cross sectional area of object measured in a plane perpendicular to

velocityD = drag coeff (0.5 for spherical object in air)

Case 2 cont.

If ΣF = mamg – R = mamg – ½ DρAv2 = m(dv/dt)

dv/dt = g – (DρA/2m)v2

Case 2 cont.

dv/dt = g – (DρA/2m)v2

Note: when a = dv/dt=0 g = (DρA/2m)v2

v = (2mg/DρA)1/2

v = terminal speed

Note: R increases with v2

R increases as ρ increases