A thin uniform rod has mass M =0.310 kg and length L= 0.450m. It has a pivot at

Question

A thin uniform rod has mass M =0.310 kg and length L= 0.450m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The rod is released from an angle theta1=65.0 degrees, and moves through its horizontal position at (B) and up to (C) where it stops with theta2=105.0 degrees, and then falls back down. Friction at the pivot is negligible. Calculate the speed of the CM at (B).

9.26 * 10^(-1)

b)The spring in (A) has a length of 0.110m and at (B) a length of 0.140m. Calculate the spring constant k.

Can someone help me do b ?

Explanation / Answer

Part B: x =x2-x1 = 0.14 -0.11 =0.03 m

From conservation of energy

K E = ELastic potential EENrgy

(1/2)Mv^2 =(1/2)kx^2

k = M[v/x]2

Need V from part A:

takn it as 7.5 m/s

K = (0.31)* (7.5/0.031)^2

k = 18145 N/m

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