Note: The pendulum bob is released at a height below the height of thepeg. Assum

Question

Note: The pendulum bob is released at a

height below the height of thepeg.

Assume: Use the small angle approximate

(sin =) tocalculate the period.

A pendulum made of a string oflength

11.3 m and a spherical bob of mass2.4kg is

able to swing in a verticalplane. The pen-

dulum is released from anangular position

49 ? from vertical as shown inthe figure be-

low. The string hits a peglocated a distance

4 m below the point ofsuspension and swings

about the peg up to anangle on the other

side of the peg. Then, the bobproceeds to

oscillate back and forth betweenthese two

angular extremities.

The acceleration of gravity is9.8m/s2 .

What is the period of the pendulumplus

peg system as shown above?Answer in units

of s.

Explanation / Answer

when the pendulum bob is released from an angular position of49o from vertical we have S1 = r1 * 1 r1 = (S1/1) S1 = 11.3 m and 1 = 49o= 49 * 0.01745 radian = 0.855 m the speed of the pendulum when it hits the string is v2 - u2 = 2g * r1 u = 0 or v = (2g * r1)1/2 --------(1) u = 0 or v = (2g * r1)1/2 --------(1) when the string hits the peg then the centripetal force actingon it is balanced by the gravitational force acting on it thereforewe get (m * v2/r2) = m * g * sin or r2 = (v2/g * sin) the value of v is obtained from equation (1),g = 9.8m/s2 and = 49o the displacement of the pendulum bob after it hits the pegis S2 = r2 * (49 + ) S2 = 11.3 - 4 = 7.3 m or = (S2/r2) - 49 let the speed of the pendulum when it goes to the oppositeend be v1 therefore we get v12 - v2 = 2g *(r1 + r2) or v1 = [v2 + 2g * (r1 +r2)]1/2 the angular momentum of the pendulum when it reaches the otherend is L = m * v1 * (r1 + r2) we know that L = I * w I = (2/5)m * r2 r is the radius of the sphere or m * v1 * (r1 + r2) =(2/5)m * r2 * w or w = [m * v1 * (r1 +r2)/(2/5)m * r2] or (2/T) = [m * v1 * (r1 +r2)/(2/5)m * r2] or T = 2 * {1/[m * v1 * (r1 +r2)/(2/5)m * r2]} m = 2.4 kg r is the radius of the sphere or m * v1 * (r1 + r2) =(2/5)m * r2 * w or w = [m * v1 * (r1 +r2)/(2/5)m * r2] or (2/T) = [m * v1 * (r1 +r2)/(2/5)m * r2] or T = 2 * {1/[m * v1 * (r1 +r2)/(2/5)m * r2]} m = 2.4 kg

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

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