The wad of clay of mass m = 0.39 kg is initially moving with a horizontal veloci

Question

The wad of clay of mass m = 0.39 kg is initially moving with a horizontal velocity v1 = 4.7 m/s when it strikes and sticks to the initially stationary uniform slender bar of mass M = 2.6 kg and length L = 1.89 m. Determine the final angular velocity of the combined body and the x-component of the linear impulse applied to the body by the pivot O during the impact. The angular velocity is positive if counterclockwise, negative if clockwise. The impulse is positive if to the right, negative if to the left.

The wad of clay of mass m = 0.39 kg is initially moving with a horizontal velocity v1 = 4.7 m/s when it strikes and sticks to the initially stationary uniform slender bar of mass M = 2.6 kg and length L = 1.89 m. Determine the final angular velocity of the combined body and the x-component of the linear impulse applied to the body by the pivot O during the impact. The angular velocity is positive if counterclockwise, negative if clockwise. The impulse is positive if to the right, negative if to the left.

Explanation / Answer

About point O angular momentum

(just before collision = just after collision )

So => mv1L/3 = I2W2

here I2 = ML^2/9 + mL^2/9

so -mv1L/3=(ML^2/9 + mL^2/9) W2

=>-mv1=L/3(M+m)W2

=>W2=-3mv1/(M+m)L

=> W2= -3*.39*4.7/{(2.6+.39)*1.89}

=> w2= -0.973 rad/sec ------------ ans(- is because of it is clockwise)

now balance of linear impulse mv1 = mw2*L/3 + Mw2*L/6 - OxdT

the x-component of the linear impulse, -OxdT= mv1 -mw2*L/3 -Mw2*L/6

=> OxdT= -mv1 + w2*L(M+2m)/6

=> OxdT= .39*4.7 - 0.973*1.89(2.6+.39)/6

=> OxdT=0.917 N.s ---------- ans(The impulse is positive)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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