Two masses (m_A = 4 kg, m_B = 10 kg) are connected via a light string around a s

Question

Two masses (m_A = 4 kg, m_B = 10 kg) are connected via a light string around a solid disk frictionless pulley (m_p = 1 kg, r_p = 20 cm). The coefficient of static and kinetic friction between all surfaces is 0.2 and 0.1 respectively. The angle theta_1 = 30 degree while theta_2 = 55 degree. (a) What is the angular acceleration of the disk? (b) What is the tension in the string on both sides of die pulley? If the system starts from rest, (c) how far does block A travel in 3 s and what is the (d) angular velocity of the disk?

Explanation / Answer

here,

mA = 4 kg

mB = 10 kg

mp = 1 kg

rp = 0.2 m

a)

the frictional force acting , ff = us * g *( mA * cos(theta1) + mB * cos(theta2) )

the accelrtion of the system , a = net force /effective mass

a = ( mB * g * sin(theta2) - mA * g * sin(theta1) - ff) / ( mA + mB + Ip /r^2)

a = ( mB * g * sin(theta2) - mA * g * sin(theta1) - us * g *( mA * cos(theta1) + mB * cos(theta2) ) ) / ( mA + mB + 0.5 * mp)

a = 9.8 * ( 10 * sin(55) - 4 * sin(30) - 0.2 *( 4 * cos(30) + 10 * cos(55) ) ) / ( 10 + 4 + 0.5 * 1)

solving for a

a = 2.94 m/s^2

teh angular accelration of the disc , alpha = a /r = 14.7 rad/s^2

b)

let the tension at A be tA

tA - mA * g * sin(theta1) - us * mA * g * cos(theta1) = mA * a

tA - 4 * 9.8 * sin(30) - 0.2 * 4 * 9.8 *cos(30) = 4 * 2.94

tA = 38.15 N

let the tension at B be tB

mB * g * sin(theta2) - tB - us * mB * g * cos(theta2) = mB * a

10 * 9.8 * sin(55) - tB - 0.2 * 10 * 9.8 *cos(55) = 10 * 2.94

tB = 39.6 N

c)

for moving system ,we take coefficient of kinetic friction

a' = net force /effective mass

a' = ( mB * g * sin(theta2) - mA * g * sin(theta1) - ff) / ( mA + mB + Ip /r^2)

a' = ( mB * g * sin(theta2) - mA * g * sin(theta1) - uk * g *( mA * cos(theta1) + mB * cos(theta2) ) ) / ( mA + mB + 0.5 * mp)

a' = 9.8 * ( 10 * sin(55) - 4 * sin(30) - 0.1 *( 4 * cos(30) + 10 * cos(55) ) ) / ( 10 + 4 + 0.5 * 1)

solving for a'

a' = 3.56 m/s^2

after t = 3 s

the distance travelled , d = 0 * t + 0.5 * a * t^2

d = 0 + 0.5 * 3.56 * 3^2

d = 16.02 m

the disatnce travelled is 16.02 m

d)

the final speed , v = 0 + a * t

v = 10.68 m/s

the angular speed of disk , w = v /r

w = 10.68/0.2 = 53.4 rad/s

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