A 4kg mass M, and radius R=.15m is released from rest at the top of an incline o

Question

A 4kg mass M, and radius R=.15m is released from rest at the top of an incline of angle 35 degrees to roll a distance of 7 meters.

A) Is there static or kinetic friction acting on the object as it rolls?

B) If a sphere and a cylinder are to start from rest and race down the incline, how far ahead in meters is the winner as it crosses the finish line at the 7.0m point?

Please help me. This is the only problem on the study exam that I do not know, and it is going to be a two question exam due to terrible exam scheduling. Thank you!

Explanation / Answer

A) We consider static friction because while the whole mass may be in motion, the part touching the ground (the only part friction can act on) is not moving relative to the ground. If it were moving, this would be considered sliding.

B) Since it is given that the object rolls a distance of 7m, we will only consider pure rolling and so the energy will be conserved.

Vertical height moves by the mass,h = 7 sin 35 = 4.02m

Decrease in gravitational potential energy = Increase in kinetic energy

or, mgh = 1/2 mv2 + 1/2 Iw2

or, 2mgh = mv2 + I(v/r)2

or, 2mgh = v2( m + I/r2)

For sphere, Isphere = 2/5 mr2 and for cylinder, Icylinder = 1/2 mr2

Velocity of the sphere at 7m mark:

2 x 4 x 9.8 x 4.02 = v2 ( m + 2/5 m)

or, 315.168 = v2 ( 28/5)

or, v = sqrt( 56.28)

or, vsphere = 7.5 m/s

Similarly, velocity of cylinder at the 7m mark:

2 x 4 x 9.8 x 4.02 = v2 ( m + 1/2 m)

or, 315.168 = v2 ( 6 )

or, v = sqrt ( 52.528)

or, vcylinder = 7.25 m/s

Again, consider this equation: 2mgh = v2( m + I/r2)

Here, h = Lsin theta

Taking derivative of both the sides:

2mgv sin theta = 2va(m+I/r2)

or, a = mgsin theta/(m+I/r2)

asphere = 4 x 9.8 x sin 35 /( 28/5)

or, asphere = 4.02 m/s2

acylinder =  4 x 9.8 x sin 35 /(6)

or, acylinder = 3.75 m/s2

Now, for the inclinde surface we have:

v= u + at

For sphere:

7.5 = 0 + 4.02 x tsphere

or, tsphere =1.87 s

For cylinder:

7.25 = 0 + 3.75 x tcylinder

or, tcylinder = 1.93 s

Thus, sphere reaches the 7m mark first in 1.87 s.

Distance covered by cylinder during this time:

s = ut + 1/2 acylindert2

or, s = 3.75 x 1.872 / 2

or, s = 6.56 m

Thus, the difference = 7 - 6.56 = 0.44 m

If a sphere and a cylinder are to start from rest and race down the incline, the sphere is ahead of the cylinder by 0.44m as it crosses the finish line at the 7.0m point.

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