spool comprised of a cylinder of radius a and negligible mass about which a ligh

Question

spool comprised of a cylinder of radius a and negligible mass about which a light cable is wound and two disks of radius b and mass m/2 on either side is on a frictionless inclined plane is illustrated D. A below. Calculate the position of the 'spool' as a function of time if it starts from rest at x-0m and t-0s. (20) E. If the cable runs out just as the 'spool' smoothly transitions to a level surface (Vem I| surface) with a coefficient of friction , under what conditions (if any) will vcm become zero (be specific). (20)

Explanation / Answer

D. given spool, cylinder of radius a, negligible mass

light cable

radii of discs = b, mass = m/2 each

position of spool as a fiunction of time = x(t)

now from force balance

let tension in the string be T

friciton = f

then from force balance

mg*sin(theta) - f - T = m*x" ( where x" = d^x/dt^2)

now,

f = mu*mg*cos(theta) ( where mu us coefficient of kinetic friction)

also, for angular acceleraiton alpha

2*0.5*(m/2)(b^2)*alpha = f*b - T*a

hence

T = [mu*mg*cos(theta)b - 0.5mb^2*alpha]/a

hence

mg*sin(theta) - mu*mg*cos(theta) - [mu*mg*cos(theta)b - 0.5mb^2*alpha]/a = m*x"

for frictionless, mu = 0

hence

g*sin(theta) + 0.5b^2*alpha/a = x"

now, alpha = theta"

if the wheel rotates by angle theta, it loses theta*a length in x direction

hence

x = theta*a

theta" = x"/a

hence

g*sin(theta) = x"(1 - b^2/2a^2)

x" = g*sin(theta)/(1 - b^2/2a^2)

integrating

x' = g*sin(theta)*t/(1 - b^2/2a^2) ( given initial speed = 0)

integrating again

x = gt^2*sin(theta)/2(1 - b^2/2a^2) ( given initial position is 0)

E. just as the cable runs out of spool

angular speed = w

linear speed = v

w = theta' = g*sin(theta)to/a(1 - b^2/2a^2)

v = x' = g*sin(theta)to/(1 - b^2/2a^2)

where to is tome after which the spool is free of string

now,

acceleration = mu*g

hence

fomr previous relations

w = v/a

assume after time t1, w becomes 0

then

final speed = U

U = v + mu*g*t1

v = mu*gt

also

from moment balance

angular deceleration = mu*g/b

hence

0 = v/a - mu*g*t1/b

hence

t1 = v*b/a*g*mu

so, U = v + mu*g*v*b/a*g*mu = v + v*b/a = v(1 + b/a)

now,

after this, theere is linear deceleration and angular acceleration

let after time t2, the linear speed = 0

hence

0 = U - mu*g*t2

t2 = U/mu*g = v(1 + b/a)/mu*g

angular speed at this instant

w = mu*g*t2/b = v(1 + b/a)/b

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