6. Roundabout 1 A ramp with length d is raised an angle θ (0◦ < θ < 90◦) above the horizontal. Ablock with mass m is placed at the top of the ramp, with the coefficient of friction between the blockand the ramp being µ. Once the block reaches the bottom of the ramp, it retains its velocity as it issmoothly transitioned onto a frictionless circular track with radius d and bank angle θ, rotating on thetrack without sliding off. A solution is a set of values {d, µ, θ} that result in the situation described above.What is the largest θ for which a solution exists?7. Roundabout 2 While the block is going around the circular track, it is given a small push perpen-dicular to its current velocity and parallel to the surface of the track, causing it to oscillate with periodT . What is the smallest possible value of T when d = 5 m?
Question
- Roundabout 1 A ramp with length d is raised an angle θ (0◦ < θ < 90◦) above the horizontal. Ablock with mass m is placed at the top of the ramp, with the coefficient of friction between the blockand the ramp being µ. Once the block reaches the bottom of the ramp, it retains its velocity as it issmoothly transitioned onto a frictionless circular track with radius d and bank angle θ, rotating on thetrack without sliding off. A solution is a set of values {d, µ, θ} that result in the situation described above.What is the largest θ for which a solution exists?7. Roundabout 2 While the block is going around the circular track, it is given a small push perpen-dicular to its current velocity and parallel to the surface of the track, causing it to oscillate with periodT . What is the smallest possible value of T when d = 5 m?
Solution
The problem is a physics problem involving concepts of friction, circular motion, and oscillation. Let's break it down into two parts:
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The block slides down the ramp under the influence of gravity and friction. The forces acting on the block are its weight (mg), the normal force (N), and the frictional force (µN). The block will slide down the ramp if the component of its weight parallel to the ramp (mg sin θ) is greater than the frictional force. So, we have mg sin θ = µN. Since the block doesn't leave the ramp, the normal force is equal to the component of the weight perpendicular to the ramp, i.e., N = mg cos θ. Substituting this into the previous equation, we get mg sin θ = µmg cos θ, which simplifies to tan θ = µ. The maximum value of θ is when µ = 1, which gives θ = 45°.
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When the block is on the circular track, it experiences a centripetal force that keeps it moving in a circle. This force is provided by the component of the weight perpendicular to the track and the normal force. The period of the oscillation is given by T = 2π √(d/g), where g is the acceleration due to gravity. Substituting d = 5 m and g = 9.8 m/s², we get T = 2π √(5/9.8) s. This is the smallest possible value of T.
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