a 15 kg mass attached to a cord that is wrapped around a pulley and released fro

Question

a 15 kg mass attached to a cord that is wrapped around a pulley and released from rest. the pulley has mass of 5 kg but is frictionless. consider the cord as mass less, inelastic and frictionless. a) determine tension in the cord? b) torque in the pulley? c) determine the acceleration of the falling mass? d)determine the time it would take the mass to fall 5 meters. e) determine angular momentum of the pulley 10 sec following the released of the falling mass from rest. f) detrmine the total kinetic energy of the system 10 seconds following the release of the falling mass from rest.

Explanation / Answer

a). assuming that pulley is like a disk.

therefore moment of inertia of the pulley. I= mR2/2

let T is the tention in the string. and acceleration of the mass= a m/s2

therefore, for falling mass,

Mg-T= Ma

or T= Mg-ma ...(1)

for the pulley,

let radius of pulley= 'R'

T*R= I

I= mR2/2 ; =a/R

therfore,

T*R= (mR2/2)*(a/R)

or T= ma/2 ....(2)

from equation (1) and (2), we can get;

Mg-ma=ma/2

Mg= 3ma/2

a=2Mg/(3m) = 2*15*9.81/(3*5) = 19.62 m/s2

therefore, tension in the string, T = ma/2 = 5*19.62/2 = 49.05 N

b).

torque in the pulley= I= (mR2/2)*(a/R) = maR/2 = 5*49.05*R/2 = 122.625*R N-m

c).

from the above calculation, u can see the acceleration of the falling mass= a = 49.05 m/s2

d).

mass is falling from the rest,

s= ut+(1/2)at2

u=0

therefore, s= (1/2)at2

5=(1/2)*49.05*t2

hence, t= 0.452 s

e).

angular momentum, L= I

I= mR2/2 ; =v/R

therefore, L = mvR/2

after 10 seconds,

velocity, v= at = 49.05*10 = 490.5 m/s

therefore, L = 5*490.5*R/2= 1226.25*R kg-m2/s

f).

total kinetic energy of the system, E= linear kinetic energy of the mass and rotational kinetic energy of the pulley = (1/2)mv2+(1/2)*I2

E = (1/2)Mv2+(1/2)*(mR2/2)(v/R)2=(1/2)Mv2 +(1/4)mv2 = (1/2)*15*490.52+(1/4)*5*490.52=2105.2 kJ

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