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Grade 12th passMechanics

A uniform solid sphere rolls on a horizontal surface at 25m/sec. If the rolls up the frictionless incline surface.
What Will be the value of height where the ball stopS? Also suggest the massAnd angle of inclined plane effect on height achieved by the ball?

Profile image of Ayesha Ghaffar
4 Years agoGrade 12th pass
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1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer0 Years ago

To determine how high a uniform solid sphere can roll up a frictionless incline after moving at a speed of 25 m/s on a horizontal surface, we can apply the principles of energy conservation. The kinetic energy of the sphere will convert into gravitational potential energy as it ascends the incline. Let’s break this down step by step.

Understanding Energy Conservation

The total mechanical energy of the sphere remains constant if we ignore air resistance and other non-conservative forces. Initially, the sphere has kinetic energy due to its motion, and as it rolls up the incline, this energy is converted into potential energy until it comes to a stop.

Kinetic Energy Calculation

The kinetic energy (KE) of a rolling sphere can be expressed as:

  • KE = (1/2)mv² + (1/2)Iω²

For a solid sphere, the moment of inertia (I) is given by:

  • I = (2/5)mr²

Since the sphere rolls without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:

  • ω = v/r

Substituting this into the kinetic energy formula gives us:

  • KE = (1/2)mv² + (1/2)(2/5)mr²(v/r)²

After simplifying, we find:

  • KE = (1/2)mv² + (1/5)mv² = (7/10)mv²

Potential Energy Calculation

The gravitational potential energy (PE) at height (h) is given by:

  • PE = mgh

Setting Up the Equation

At the point where the sphere stops rolling up the incline, all its kinetic energy has been converted into potential energy:

  • (7/10)mv² = mgh

Notice that the mass (m) cancels out, simplifying our equation to:

  • (7/10)v² = gh

Solving for Height

Now, we can solve for height (h):

  • h = (7/10)(v²/g)

Substituting the values, where v = 25 m/s and g = 9.81 m/s²:

  • h = (7/10)(25²/9.81)
  • h ≈ (7/10)(625/9.81) ≈ (7/10)(63.7) ≈ 44.6 m

Influence of Mass and Angle

The mass of the sphere does not affect the height achieved, as seen in our calculations where it cancels out. However, the angle of the incline can influence how quickly the sphere reaches that height. A steeper incline would mean a quicker conversion of kinetic energy to potential energy, but the maximum height achieved remains the same as long as the incline is frictionless.

In summary, the sphere will roll up to a height of approximately 44.6 meters on a frictionless incline after starting at 25 m/s. The mass of the sphere does not affect this height, while the angle of the incline affects the rate of ascent but not the maximum height reached.