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Q103P

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Fundamentals Of Physics
Found in: Page 294

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Short Answer

In Fig10-63., a thin uniform rod (mass, 3.0  kglength4.0  m) rotates freely about a horizontal axis A that is perpendicular to the rod and passes through a point at distanced=1.0 m from the end of the rod. The kinetic energy of the rod as it passes through the vertical position is20 J. (a) What is the rotational inertia of the rod about axis A? (b) What is the (linear) speed of the end B of the rod as the rod passes through the vertical position? (c) At what angle u will the rod momentarily stop in its upward swing?

a) Rotational inertia (M.I) of the rod about axis A is. 7.0 kg.m2

b) Linear speed of point B of the rod when the rod is vertical is 7.2  m/s.

c)The angle at which the rod momentarily stops in its upward swing is equal t 71.35°o.

See the step by step solution

Step by Step Solution

Step 1: Given

i) Length of the rod is, L=4.0 m.

ii) Mass of the rod is, M=3.0 kg. M=3.0 kg.

iii) Distance,d=1.0  m.

iv)The K.E. of the rod as when it is vertical is, K.E=20 J.

Step 2: Determining the concept

Find the moment of inertia of the rod about an axis A using the parallel axis theorem. Using the formula for rotational K.E., find the angular speed of the rod. Then, using it’s relation with linear speed, find the linear speed of the end B of the rod as it passes through the vertical position. Using energy conservation, find the angle at which the rod momentarily stops in its upward swing.

The parallel axis theorem is given as-

I=Icom+MR2

M.I. ofa uniform rod about its center of mass is,

role="math" localid="1660999291388" I=112ML2

Kinetic energy is given as-

K.Erot=12I ω2

where, Iis moment of inertia, m, M are masses, r is radius, v is velocity, is angular frequency, g is acceleration due to gravity, his height, L is length, K.E. is kinetic energy.

Step 3: (a) Determining the rotational inertia (M.I) of the rod about axis A

According to the parallel axis theorem,

I=Icom+MR2

Applying to the given system, the M.I of rod about axis A is,

I=Icom+Md2

I=112ML2+Md2

I=112(3 kg)(4 m)2+(3 kg)(1 m)2

I=7.0  kg.m2

Therefore, rotational inertia (M.I.) of the rod about axis A is 7.0 kg.m2.

Step 4: (b) Determining the linear speed of the end B of the rod as it passes through the vertical position

The rotational K.E of the rod is,

K.Erot=12I ω2

ω2=2K.ErotI

ω=2K.ErotI

ω=2(20 J)7 kgm2

ω=2.4 rad/s

But,v=rω.  Hence, linear velocity of rod is,

v=(3 m)(2.4 rad/s)

v=7.2  m/s

Therefore, linear speed of the end B of the rod as it passes through the vertical position is. 7.2  m/s

Step 4: (b) Determining the angle at which the rod momentarily stops in its upward swing 

As per the law of conservation of energy, when the rod momentarily stops it has zero K.E and maximum P.E. Using analogy with simple pendulum, write for maximum P.E as,

P.E=mgh=mgd(1cosθ)

But, P.E=K.Erot

20 J=(3​ kg)(9.8 m/s2)(1 m)(1cosθ)

(1cosθ)=0.68

cosθ=0.32

θ=71.35°

Hence, the angle at which the rod momentarily stops in its upward swing is equal to.71.35°

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