1. (33 points) The figure below shows two blocks of mass mi = 250g and m2 = 400g

Question

1. (33 points) The figure below shows two blocks of mass mi = 250g and m2 = 400g are connected by a massless cord wrapped around a uniform disk of mass M = 740g and radius R = 11.0cm. Assume the disk rotates without friction about a horizontal axis through its center and that the cord does not slip on the disk. When the blocks are released from rest find the tension in the left cord (T1), the tension in the right cord (T2), the magnitude of the angular acceleration of the disk, and the magnitude of the (linear) acceleration of the blocks.

Explanation / Answer

let a is the linear acceleration of m1 and m2.

net force on m1,

Fnet1 = T1 - m1*g

m1*a = T1 - m1*g

==> T1 = m1a + m1*g ---(1)

net force on m2,

Fnet2 = m2*g - T2

m2*a = m2*g - T2

==> T2 = m2*g - m2*a --(2)

Net torque acting on disk

T2*R - T1*R = I*alfa

T2*R - T1*R = I*a/R (since a = R*alfa)

T2 - T1 = I*a/R^2

T2 - T1 = (0.5*M*R^2)*a/R^2

T2 - T1 = 0.5*M*a

(m2*g - m2*a) -(m1a + m1*g) = 0.5*M*a

(m2-m1)*g = (0.5*M + m1+m2)*a

==> a = (m2-m1)*g/(0.5*M + m1+m2)

= (0.4-0.25)9.8/(0.5*0.74 + 0.25 + 0.4)

= 1.44 m/s^2 <<<<<<<<<<<<---------------------------Answer

from equation 1

T1 = m1*(g+a)

= 0.25*(9.8+1.44)

= 2.81 N <<<<<<<<<<<<---------------------------Answer

from equation 2

T2 = m2*(g-a)

= 0.4*(9.8-1.44)

= 3.34 N <<<<<<<<<<<<---------------------------Answer

angular accelartion, alfa = a/R

= 1.44/0.11

= 13.1 rad/s^2 <<<<<<<<<<<<---------------------------Answer

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.