A solid sphere is rolling on a frictionless s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the solid sphere to climb the inclined surface to a height $h$,its total initial kinetic energy must be greater than or equal to the potential

Vedclass · Accepted Answer

$v \ge \sqrt {\frac{10}{7}gh}$

Vedclass · Answer

(A) For the solid sphere to climb the inclined surface to a height $h$,its total initial kinetic energy must be greater than or equal to the potential energy gained at height $h$.The total kinetic energy $(K.E.)_{total}$ is the sum of translational kinetic energy $(K.E.)_{T}$ and rotational kinetic energy $(K.E.)_{R}$:$(K.E.)_{total} = (K.E.)_{T} + (K.E.)_{R} \ge P.E.$$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \ge mgh$For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.Substituting these values:$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 \ge mgh$$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 \ge mgh$$\frac{7}{10}mv^2 \ge mgh$$v^2 \ge \frac{10gh}{7}$$v \ge \sqrt{\frac{10gh}{7}}$

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

$A$ solid sphere is rolling on a frictionless surface with a translational velocity $v \ m/s$,as shown in the figure. If it is to climb the inclined surface to a height $h$,then $v$ should be:

Similar Questions

The rate of the reaction $2N_2O_5 \to 4NO_2 + O_2$ can be written in three ways:
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$
$\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

The general solution of the equation $(\sqrt{3} - 1)\sin \theta + (\sqrt{3} + 1)\cos \theta = 2$ is

The $\Delta H^o$ of combustion of $C_4H_{10(g)}$ is $-2650 \ kJ \ mol^{-1}$. The amount of heat evolved at constant pressure that can be obtained by burning $5.8 \ kg$ of $C_4H_{10}$ is approximately:

Which of the following statements is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module