Your submissions: | 5.8/ Computed value: 5.8 Submitted: Monday, October 30 at 2:

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Your submissions: | 5.8/ Computed value: 5.8 Submitted: Monday, October 30 at 2:19 PM Feedback: Correct! 5) What is the acceleration of the center of mass of the meterstick when it is vertical? m/s2 Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 6) What is the tension in the string when the meterstick is vertical? Submit You currently have O submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 7) Where is the angular accelerbtion of the meterstick a maximum? right after the string is cut and the meterstick is still horizontal when the meterstick is vertical- at the bottom of its path the angular acceleration is constant Submit You currently have 3 submissions for this question. Only 10 submission are allowed. You can make 7 more submissions for this question. (Survey Question) 8) Below is some space to write notes on this problem

Explanation / Answer

given, length of meter stick, l = 1 m

mass,m = 0.231 kg

it hangs from two short sticks at 0.25 m and 0.75 m

1. let tnesion in both the strings be T ( from symmetry)

then from force balance

2T = mg

T = mg/2 = 0.231*9.81/2 = 1.133055 N

2. when the right string is cut, the pivot point is the 0.25 m mark ( left stirng)

initial angular acceleration = alpha

I*alpha = mg*0.25

now I = ml^2/12 = m*0.25^2 = 0.231(1/12 + 0.25^2) = 0.0336875 kg m^2

hence

alpha = 16.8171 rad/s/s

3. after the rigth string is cut tension in the left string = T'

initial linear acceleration of COM = alpha*0.25 = 4.2042 m/s/s

hence from force balance

mg - T' = m*4.20428

T' = m(g - 4.204285) = 1.29492 N

4. when the meter stick becomes vertical let the speed of the center of mass be v

then

initial PE = final KE

m*g*0.25 = 0.5mv^2

v^2 = 0.5g

v = 2.2147 m/s

hence acceleration of COM = v^2/0.25 = 19.62 m/s/s

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