NEB Class 11 • Physics • Worked Practice

Circular Motion Practice Set: NEB 11 Physics

Build from angular conversions to radial force equations, road curves and vertical circles. Every solution includes a direction, assumption or dimensional check.

Fifteen graded worked questions
Concept, calculation and derivation practice
Timed mixed challenge and error diagnosis

Velocity is tangential; radial acceleration points inward.

Practice boundary

Skills measured in this set

The set covers period, frequency and angular speed; v = rω; radial acceleration; inward force models; level and banked curves; and vertical-circle equations. Review the concept guide or study guide if a method is unfamiliar. The CDC Physics Grade 11 page remains the Nepal scope checkpoint.

Solutions use the direction conventions and equations described in the OpenStax circular-motion section and the real-force interpretation from its centripetal-force section.

Level 1

Angular and linear variables

1. Revolutions per minute

A wheel rotates at 240 rpm. Find frequency, period and angular speed.

Solution: f = 240/60 = 4.0 Hz; T = 1/f = 0.25 s; ω = 2πf = 8π ≈ 25.1 rad s⁻¹.

2. Tangential speed

The wheel radius is 0.30 m at 4.0 Hz. Find rim speed.

Solution: v = rω = 0.30×8π ≈ 7.54 m s⁻¹. A point halfway out has half this speed but the same ω.

3. Number of turns

A disc turns at 5 Hz for 12 s. Find revolutions and angular displacement.

Solution: N = ft = 60 revolutions; θ = 2πN = 120π rad.

Level 2

Radial acceleration

4. Speed and radius

A particle moves at 9.0 m s⁻¹ on a 3.0 m radius circle. Find radial acceleration.

Solution: a_r = v²/r = 81/3 = 27 m s⁻² inward.

5. Angular-speed form

A point 0.20 m from an axis rotates at 10 rad s⁻¹.

Solution: a_r = rω² = 0.20×100 = 20 m s⁻². Unit: m×s⁻².

6. Scaling

Speed doubles and radius triples. How does a_r change?

Solution: a′/a = (2²)/3 = 4/3. It increases by one-third, not by a factor of two.

7. Total acceleration

At an instant, a_r = 12 m s⁻² and tangential acceleration is 5 m s⁻².

Solution: The components are perpendicular, so a = √(12²+5²) = 13 m s⁻².

Level 3

Real forces providing inward acceleration

8. Horizontal string

A 0.40 kg ball moves at 5.0 m s⁻¹ in a horizontal circle of radius 2.0 m. Find inward net force.

Solution: F_in = mv²/r = 0.40×25/2 = 5.0 N inward. If horizontal tension is the only radial force, tension is 5.0 N.

9. Conical pendulum components

A mass moves in a horizontal circle while its string makes angle θ with vertical.

Solution: Vertical balance is T cosθ = mg; horizontal inward equation is T sinθ = mv²/r. Dividing gives tanθ = v²/(rg).

10. Release direction

A string breaks while a stone moves counter-clockwise at the rightmost point of its circle. What is the initial path?

Solution: The stone initially moves vertically upward, tangent to the circle. It does not move radially outward.

Level 4

Level and banked curves

11. Level curve limit

For r = 40 m and μ_s = 0.50, estimate the ideal maximum speed using g = 9.8.

Solution: μmg = mv²/r gives v = √(μrg) = √(0.50×40×9.8) = 14 m s⁻¹. Mass cancels. The result belongs to a simplified limiting model.

12. Ideal bank angle

A 75 m radius road is ideally banked for 15 m s⁻¹.

Solution: tanβ = v²/(rg) = 225/(75×9.8) ≈ 0.306, so β ≈ 17.0°. Friction is neglected.

13. Design comparison

At the same radius, design speed rises from 10 to 20 m s⁻¹. Compare tanβ.

Solution: It scales with v², so tanβ becomes four times larger.

Level 5

Vertical-circle force and energy

14. Tension at the bottom

A 0.50 kg ball moves at 6.0 m s⁻¹ at the bottom of a 1.2 m vertical circle.

Solution: Inward is upward: T − mg = mv²/r. Thus T = 0.50×36/1.2 + 4.9 = 19.9 N.

15. Tension at the top

The same mass moves at 5.0 m s⁻¹ at the top.

Solution: Inward is downward: T + mg = mv²/r, so T = 0.50×25/1.2 − 4.9 ≈ 5.52 N. Positive tension means the string can remain taut.

16. Limiting top speed

For radius 0.80 m, find the minimum top speed for a just-taut string.

Solution: At the limit T = 0, so mg = mv²/r and v = √(gr) = √(9.8×0.80) ≈ 2.80 m s⁻¹.

17. Bottom speed for just-complete circle

Using the limiting top speed and energy conservation, find the minimum bottom speed.

Solution: ½mv_b² = ½m(gr)+mg(2r). Hence v_b² = 5gr and v_b = √(5gr). For r = 0.80 m, v_b ≈ 6.26 m s⁻¹.

Level 6

Multi-stage transfer problems

18. Rotating platform comparison

Points A and B are 0.20 m and 0.50 m from the axis of a platform rotating at 3.0 rad s⁻¹. Compare their speeds and radial accelerations.

Solution: v_A = 0.60 m s⁻¹, v_B = 1.50 m s⁻¹. Their accelerations are rω²: 1.8 and 4.5 m s⁻². Both ratios equal r_B/r_A = 2.5 because ω is shared.

19. Two-speed road comparison

A vehicle negotiates the same level curve first at v and later at 1.5v. Compare the required friction.

Solution: Required inward force scales with v², so it becomes 1.5² = 2.25 times as large. This shows why modest speed increases can strongly raise the friction demand.

20. Vertical-circle energy and force

A 1.0 kg mass on a 1.0 m string moves at 7.0 m s⁻¹ at the bottom. Neglect resistance. Find its top speed and top tension.

Solution: Energy gives ½×7² = ½v_t² + g(2), so v_t² = 49−39.2 = 9.8 and v_t ≈ 3.13 m s⁻¹. At the top, T+mg = mv_t²/r, so T = 9.8−9.8 = 0 N. The string is just taut in the ideal model.

These questions require two decisions, not merely two formulas. Mark where the system, position or fixed quantity changes before starting algebra.

18-minute mixed challenge

Questions without method labels

A 0.25 m wheel turns at 600 rpm. Find rim speed and radial acceleration.
A 900 kg vehicle turns at 12 m s⁻¹ on a 60 m level curve. Find required friction.
Explain why static rather than kinetic friction normally turns a rolling tyre.
Derive tanβ = v²/(rg) for an ideal bank.
Draw force diagrams at four positions of a vertical circle.
At what top speed does a 2 m string just remain taut?

Checks: (1) 5π m s⁻¹, about 987 m s⁻²; (2) 2160 N inward; (6) √19.6 ≈ 4.43 m s⁻¹.

Use the PhET Gravity and Orbits simulation for a predict–observe–explain extension: change speed or central mass and describe curvature before running it.

Error analysis

Common mistakes and exact repairs

rpm error

Divide by 60 before treating rpm as hertz.

Direction error

Draw the tangent and inward radius before equations.

Force error

Do not add an extra centripetal force to real forces.

Position error

Rewrite the radial equation at each vertical-circle position.

For online or physical NEB tuition, call 9846662070 with the first wrong decision from your attempt.

Frequently asked questions

Questions about the Circular Motion practice set

How should I mark a circular-motion solution?

Credit the diagram, conversions, inward direction, real-force equation, substitution, units and a physical check.

Why did mass cancel in the road-speed formula?

Both limiting friction μmg and required radial force mv²/r are proportional to mass in the ideal model.

Can radial acceleration change speed?

Radial acceleration alone is perpendicular to velocity and changes direction, not speed. Tangential acceleration changes speed.

Why is tension different at the top and bottom?

Gravity contributes differently to the inward force, and speed may also differ because height changes.

What score shows readiness?

Aim for at least 80% on an unseen mixed set with no repeated direction or double-counting error.

Where can I get help with NEB Circular Motion?

For current online or physical tuition options, call 9846662070 and confirm timetable, class mode, teacher availability and fees.

References and next steps

Sources and related study guides

Use the Physics revision roadmap to schedule mixed Mechanics retrieval. Scope and sources were checked on 2 August 2026. Follow current CDC, NEB and college instructions if requirements change.

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