HI GUYS CAN YOU SOLVE THIS ASAP PLS AND THANKS! QUESTION 1.I Consider a rigid bo

Question

HI GUYS CAN YOU SOLVE THIS ASAP PLS AND THANKS!

QUESTION 1.I

Consider a rigid body that is rotating. Which of the following is an accurate statement?

All points on the body are moving with the same linear velocity.

All points on the body are moving with the same angular velocity.

Its center of rotation is accelerating.

Its center of rotation is its center of gravity.

Its center of rotation is at rest, i.e., not moving.

QUESTION 1.ii

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm mark. What is the mass of the meter stick?

178 g

73.4 g

32.6 g

127 g

89.7 g

All points on the body are moving with the same linear velocity.

All points on the body are moving with the same angular velocity.

Its center of rotation is accelerating.

Its center of rotation is its center of gravity.

Its center of rotation is at rest, i.e., not moving.

Explanation / Answer

1.

Since for a rigid body point of rotation is fixed, So All points on the body are moving w.r.t that point. Which means Angular velocity of all the points is same.

Mathematically

w = v/r

Since r = distance from point of rotation

Correct option is 2.

Part 2.

Using torque balance on meter stick,

Since Net torque is zero, So

Net Torque = 0

Clockwise torque = Anti-clockwise torque

r1*F1 = r2*F2

r1*m1*g = r2*m2*g

r1*m1 = r2*m2

r1 = distacne of mass placed from balanced point = 90 cm - 61.3 cm = 28.7 cm

m1 = mass placed = 50 gm

r2 = distance between original center of mass and after putting the new mass = 61.3 - 50 = 11.3 cm

So,

m2 = mass of meterstick = m1*r1/r2

m2 = 50*28.7/11.3

m2 = 126.99 = 127 gm

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