A meterstick pivoted about a horizontal axis through the O-cm end is held in a horizontal position and let go. (a) \A{rat is the tangential acceleration of the 100-cm position? Are you surprised by this result? (b) Which position has a tangential acceleration equal to the acceleration due to gravity?

A. Tangential velocity of the 100cm position = 1.5g

B. Tangential velocity at 66.67cm equals acceleration due to gravity.

Explanation:

I = I(cm) × M.(1/2)²

I = 1/12.m.(1)² + m(1/2)²

= 1/3m

= 1/2.mg = mg/2

Therefore, ∆= I׶

¶ = ∆/I = (mg/2)/(mg/3) = 3g/2

A. Tangential acceleration of the 100cm position = ∆ × 1

= 1×3g/2 = 1.5g

B. At g = ¶ × 3g/2

¶ = 2/3m

= 2/3×100

= 200/3

Hence, tangential acceleration is equal to = 66.67cm

none, its a machine

answer: b.

explanation: