---
title: "Radial and transverse force on block B"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:block-b-polar-radial-transverse-force"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-b-polar-radial-transverse-force
site: BCT Engineering
---

# Radial and transverse force on block B

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2073 Shrawan, 8+2 marks
- Video: [Radial & Transverse Force | Polar Acceleration | Circular Dynamics](https://www.youtube.com/watch?v=6uVGIog84m8) — Engineering Hack
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-b-polar-radial-transverse-force
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

The motion of a 1000 gm block B in a horizontal plane is defined by radius, r = 2(1+cos 2πt) and θ = 2πt where r is expressed in meters and t in seconds. Determine the radial and transverse components of the force exerted on the block B at 0.8 sec. Explain about principle of impulse and momentum.

## Short answer

At t=0.8 s: r=2.618 m, ṙ=11.95 m/s, r̈=−24.40 m/s², θ̇=6.283 rad/s. a_r = –127.75 m/s² → F_r = –127.75 N; a_θ = 150.18 m/s² → F_θ = 150.18 N. Impulse-momentum: the net impulse equals change in momentum, ∫F dt = mΔv.

## Answer

#### Data
m = 1000 g = 1 kg; r = 2(1 + cos 2πt) m; θ = 2πt rad. At t = 0.8 s: 2πt = 5.0265 rad.
- cos 2πt = 0.3090, sin 2πt = −0.9511
- r = 2(1.3090) = 2.618 m
- ṙ = −4π sin 2πt = −4π(−0.9511) = 11.95 m/s
- r̈ = −8π² cos 2πt = −78.96(0.3090) = −24.40 m/s²
- θ̇ = 2π = 6.283 rad/s, θ̈ = 0

#### Acceleration components
- a_r = r̈ − rθ̇² = −24.40 − 2.618(39.48) = −127.75 m/s²
- a_θ = rθ̈ + 2ṙθ̇ = 0 + 2(11.95)(6.283) = 150.18 m/s²

#### Force components (horizontal plane, so no weight effect)
- F_r = m a_r = 1 × (−127.75) = **−127.75 N** (directed toward O)
- F_θ = m a_θ = 1 × 150.18 = **150.18 N** (transverse, in the direction of increasing θ)

Resultant force = √(127.75² + 150.18²) = 197.2 N.

#### Principle of impulse and momentum
Starting from Newton's second law, F = m dv/dt, multiply by dt and integrate between times t₁ and t₂:

∫ F dt = m v₂ − m v₁

- **Linear impulse** = ∫ F dt (units N·s): the effect of a force acting over a time interval.
- **Linear momentum** = m v (units kg·m/s).
- The principle states: *the linear impulse of the resultant force on a particle during a time interval equals the change in its linear momentum.* In vector form m v₁ + ∫ F dt = m v₂, with separate component equations along x, y, z.
- If the resultant force is zero, momentum is conserved. The principle is most useful when the force is given as a function of time or when velocity and time (not displacement) are involved, for example impact.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which is the correct radial acceleration in polar coordinates?
   A. a_r = r_ddot + r theta_dot^2
   B. a_r = r_ddot - r theta_dot^2
   C. a_r = r theta_ddot + 2 r_dot theta_dot
   D. a_r = r_dot - r theta_dot
   Answer: B. a_r = r_ddot - r theta_dot^2. The second term is the centripetal part and gives -127.75 m/s^2 here. (Key point: Radial acceleration formula)

2. Which is the correct transverse acceleration?
   A. a_theta = r theta_ddot + 2 r_dot theta_dot
   B. a_theta = r theta_ddot - 2 r_dot theta_dot
   C. a_theta = r_ddot + r theta_dot^2
   D. a_theta = r theta_dot^2
   Answer: A. a_theta = r theta_ddot + 2 r_dot theta_dot. With theta_ddot = 0 only the Coriolis term 2 r_dot theta_dot = 150.18 m/s^2 remains. (Key point: Transverse acceleration formula)

3. Why does the weight of the block not appear in the force components?
   A. The block is light
   B. The motion is in a horizontal plane, so weight acts perpendicular to it and F = m a applies in-plane
   C. Weight always cancels radially
   D. Gravity is zero in polar coordinates
   Answer: B. Weight is balanced by the plane's reaction, so only in-plane forces equal m a_r and m a_theta, with m = 1 kg. (Key point: Force from acceleration)

4. The block is given as 1000 gm. What mass is used in F = m a?
   A. 1000 kg
   B. 100 kg
   C. 1 kg
   D. 9.81 kg
   Answer: C. 1000 g = 1 kg, so F_r = -127.75 N and F_theta = 150.18 N. (Key point: Mass unit conversion)

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## Related in other subjects

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