7) A slender unifrom rod 100.00 cm long is used as a meter stick. Two parallel a

Question

7) A slender unifrom rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considered. The first axis passes through the 50-cm mark and the second axis passes through the 30cm mark. What is the ratio of the moment of inertia through the second axis to the moment of inertia through the first axis?

A) I2/I1=2.3

B) I2/I1= 1.9

C) I2/I1= 2.1

D) I2/I1 =1.5

E) I2/I1=1.7

8) A 95 N force exerted at the end of a 0.50 m long torque wrench givees rise to a torquw of 15 N m. What is the angle ( assumed to be less than 90) between the wrench handle and the direction of the applied force?

A) 14 B) 18 C) 25 D) 22

9) A turbine blade rotates with angular velocity w(t) = 2.00 rad/s - 2.10 rad/ s3t2. how many revolutions has the blade made at t= 7.3s?

A) 68 B) 41 C) 27 D) 53 E)76

10) While spining down from 300 rpm to rest, a solid uniform flywheel does 7.7 kJ of work. If the radius of the is 0.9 m, what is its mass?

A) 27.8 Kg B) 0.152 Kg C) 38.5 Kg D) 12.7 Kg

Explanation / Answer

7) Moment of inertia about axis passing through centre = mr^2 / 12

so I1 = Ic = mr^2 / 12

distance between 50cm and 30 cm mark = 0.2r  

using parrallel axis theorem ,

I2 = mr^2 / 12 + m(0.2r)^2 = 1.48 mr^2 / 12

so I2 / I2 = 1.48 ans

8) torque = r x F = rFsin@

15 = 0.50 x 95 x sin@

sin@ = 0.316

@ = 18.40 degrees

9) w = 2 - 2.10t^2

theta = intergral of w.dt = 2t - 2.10t^3 /3 = 2t - 0.7t^3

@(t) = 2t - 0.7t^3

@(7.3) = 257.71 rad

revolitions = 257.71 / 2 xpi   = 41 rev

10) Work done = change in K.E.

7.7 x 10^3 = Iw^2 /2

7700 = (m x 0.9^2 /2 ) x (300 x 2pi /60)^2 /2  

m =38.5 kg

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