A thin uniform rod (mass = 2.0kg, length = 0.70m) is free torotate about a frict

Question

A thin uniform rod (mass = 2.0kg, length = 0.70m) is free torotate about a frictionless pivot at one end. the rod is releasedfrom rest in the horizontal position. What is the magnitude of theangular acceleration of the rod at the instant it is 50 degreesbelow the horizontal? (g = 9.8 m/s2)
A.) 7 rad/s2
B.) 9 rad/s2
C.) 13 rad/s2
D.) 5 rad/s2
E.) 15 rad/s2 A thin uniform rod (mass = 2.0kg, length = 0.70m) is free torotate about a frictionless pivot at one end. the rod is releasedfrom rest in the horizontal position. What is the magnitude of theangular acceleration of the rod at the instant it is 50 degreesbelow the horizontal? (g = 9.8 m/s2) A.) 7 rad/s2
B.) 9 rad/s2
C.) 13 rad/s2
D.) 5 rad/s2
E.) 15 rad/s2 B.) 9 rad/s2 C.) 13 rad/s2 D.) 5 rad/s2 E.) 15 rad/s2

Explanation / Answer

We know that S = r * where r = 0.70 m and = 50o = 50 * 0.0349radians = 1.74 radians or S = 0.70 * 1.74 = 1.22 m We know from the relation S = ut + (1/2)gt2 where u = 0 and g = 9.8 m/s2 or 1.22 = 0 * t + (1/2) * 9.8 * t2 or t2 = (1.22/4.9) or t = 0.498 s We know that v = u + gt or v = 0 + 9.8 * 0.498 = 4.9 m/s The linear velocity is v = r * w or w = (v/r) = (4.9/0.70) = 7 rad/s Let the angular acceleration of the rod be therefore weget w = wo + t or = (w - wo/t) = (7 - 0/0.498) = 14rad/s2 or t2 = (1.22/4.9) or t = 0.498 s We know that v = u + gt or v = 0 + 9.8 * 0.498 = 4.9 m/s The linear velocity is v = r * w or w = (v/r) = (4.9/0.70) = 7 rad/s Let the angular acceleration of the rod be therefore weget w = wo + t or = (w - wo/t) = (7 - 0/0.498) = 14rad/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---