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A solid disk of radius 8.10 cm and mass 1.55 kg, which is rolling at a speed of 2.40 m/s, begins rolling without slipping up a 15.0 degree slope. How long will it take for the disk to come to a stop?

2 Answers

2 votes

Final answer:

To find the time it takes for the disk to come to a stop, we can use the principle of conservation of energy. The initial mechanical energy of the system is equal to the sum of the kinetic energy and the gravitational potential energy at the starting position. The final time it takes for the disk to come to a stop is approximately 0.3466 seconds.

Step-by-step explanation:

To find the time it takes for the disk to come to a stop, we can use the principle of conservation of energy. The initial mechanical energy of the system is equal to the sum of the kinetic energy and the gravitational potential energy at the starting position. As the disk rolls up the slope, its mechanical energy is gradually converted into rotational kinetic energy and gravitational potential energy. When the disk comes to a stop, all of its initial mechanical energy is converted into potential energy at the highest point of the slope.

The formula for the gravitational potential energy is: PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, since the disk starts rolling up the slope from rest, its initial kinetic energy is zero. Therefore, the initial mechanical energy is equal to the gravitational potential energy at the starting position:

ME_initial = m * g * h

At the highest point of the slope, the disk comes to a stop, so its final kinetic energy is zero. Therefore, the final mechanical energy is equal to the gravitational potential energy at the highest point of the slope:

ME_final = m * g * h'

Since energy is conserved, we can equate the initial and final mechanical energies:

m * g * h = m * g * h'

From this equation, we can solve for h' (the height at which the disk comes to a stop) by rearranging the equation:

h' = h * (m * g) / (m * g)

Plugging in the given values, we have:

h' = 15.0 degrees * (1.55 kg * 9.8 m/s^2) / (1.55 kg * 9.8 m/s^2)

h' = 15.0 degrees

Therefore, the time it takes for the disk to come to a stop is given by:

t = 0 - 2.40 m/s / a, where a is the acceleration of the disk up the slope. The acceleration can be determined using the components of the gravitational force that act parallel and perpendicular to the slope. The parallel component is given by:

F_parallel = m * g * sin(theta), where theta is the angle of the slope. The perpendicular component is given by:

F_perpendicular = m * g * cos(theta)

The net force acting on the disk up the slope is equal to the difference between the parallel and perpendicular components:

F_net = F_parallel - F_perpendicular

Using Newton's second law of motion, we can relate the net force with the acceleration:

F_net = m * a

Therefore, we can solve for the acceleration:

a = (m * g * sin(theta) - m * g * cos(theta)) / m

a = g * (sin(theta) - cos(theta))

Plugging in the given values, we have:

a = 9.8 m/s² * (sin(15.0 degrees) - cos(15.0 degrees))

a = 9.8 m/s² * (0.2588 - 0.9659)

a = 9.8 m/s² * (-0.7071)

a = -6.9431 m/s²

Since the acceleration is negative, it indicates that the disk is decelerating up the slope. Finally, we can calculate the time it takes for the disk to come to a stop:

t = 2.40 m/s / 6.9431 m/s²

t = 0.3466 s

User SavageWays
by
8.2k points
2 votes

The time taken for the disk to come to a stop is 6.96 seconds.

How to calculate the time taken for the disc to come to stop?

The time taken for the disk to come to a stop is calculated by applying the following formula;

P.E = K.E

mgh = ¹/₂mv² + ¹/₂Iω²

mgLsin(θ) = ¹/₂mv² + ¹/₂Iω²

where;

  • I is the moment of inertia of the disk
  • L is the length of the incline
  • ω is the angular speed of the disk

mgLsin(θ) = ¹/₂mv² + ¹/₂Iω²

mgLsin(θ) = ¹/₂mv² + ¹/₂(¹/₂mr²)ω²

mgLsin(θ) = ¹/₂mv² + ¹/₄mr²ω²

mgLsin(θ) = ¹/₂mv² + ¹/₄mr²(v/r)²

mgLsin(θ) = ¹/₂mv² + ¹/₄mr²(v²/r²)

mgLsin(θ) = ¹/₂mv² + ¹/₄mv²

mgLsin(θ) = ³/₄mv²

gLsin(θ) = ³/₄v²

L = (3v²) / ( 4 x sin(θ) )

The distance traveled by the disk is calculated as;

L = ( 3 x 2.4²) / ( 4 x sin (15) )

L = 16.7 m

The time of motion of the disk is;

t = L / v

t = 16.7 m / 2.4 m/s

t = 6.96 s

User Abdullah Qudeer
by
7.6k points
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