Join WhatsApp Icon JEE WhatsApp Group
Question 5

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

$$I_P = I_{\text{cm}} + MR^2 = 2MR^2$$

$$\tau_P = F \cdot (2R)$$

$$\tau_P = I_P \alpha \implies 2FR = 2MR^2 \alpha$$

$$\alpha = \frac{F}{MR} = \frac{40}{5 \times 0.5} = 16\text{ rad/s}^2$$

Get AI Help

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests
Ask AI